Calculate the value of \[\dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\].
Answer
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Hint: Here, we need to simplify the given expression. We will convert the decimal numbers in the fraction form such that the denominator is a multiple of 10. Then we will simplify the given expression using rules of exponents to get the required value.
Formula Used:
We will use the following formulas:
1.\[{a^b} \times {a^c} = {a^{b + c}}\]
2.\[{a^m} \times {b^m} = {\left( {ab} \right)^m}\]
3.\[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\]
4.\[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\].
Complete step-by-step answer:
We will rewrite the expression \[\dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\] to find the required value.
Rewriting the decimal numbers as fractions, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{\dfrac{{6626}}{{1000}} \times {{10}^{ - 34}}}}{{\sqrt {2\left( {\dfrac{{91}}{{10}} \times {{10}^{ - 31}}} \right)\left( {\dfrac{{16}}{{10}} \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\]
Rewriting 50 as the product of 5 and 10, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{\dfrac{{6626}}{{1000}} \times {{10}^{ - 34}}}}{{\sqrt {2\left( {\dfrac{{91}}{{10}} \times {{10}^{ - 31}}} \right)\left( {\dfrac{{16}}{{10}} \times {{10}^{ - 19}}} \right)\left( {5 \times 10 \times {{10}^3}} \right)} }}\]
Rewriting 10 as 10 raised to the power 1, and 1000 as 10 raised to the power 3, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{\dfrac{{6626}}{{{{10}^3}}} \times {{10}^{ - 34}}}}{{\sqrt {2\left( {\dfrac{{91}}{{{{10}^1}}} \times {{10}^{ - 31}}} \right)\left( {\dfrac{{16}}{{{{10}^1}}} \times {{10}^{ - 19}}} \right)\left( {5 \times {{10}^1} \times {{10}^3}} \right)} }}\]
If two numbers with same base and different exponents are multiplied, the product can be written as \[{a^b} \times {a^c} = {a^{b + c}}\].
If two or more numbers with different bases and same exponent are divided, the quotient can be written as \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\].
Therefore, rewriting the expression using the rules of exponents \[{a^b} \times {a^c} = {a^{b + c}}\] and \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times \dfrac{{{{10}^{ - 34}}}}{{{{10}^3}}}}}{{\sqrt {2\left( {91 \times \dfrac{{{{10}^{ - 31}}}}{{{{10}^1}}}} \right)\left( {16 \times \dfrac{{{{10}^{ - 19}}}}{{{{10}^1}}}} \right)\left( {5 \times {{10}^{1 + 3}}} \right)} }}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 34 - 3}}}}{{\sqrt {2\left( {91 \times {{10}^{ - 31 - 1}}} \right)\left( {16 \times {{10}^{ - 19 - 1}}} \right)\left( {5 \times {{10}^{1 + 3}}} \right)} }}\end{array}\]
Adding the subtracting the terms in the exponents, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {2\left( {91 \times {{10}^{ - 32}}} \right)\left( {16 \times {{10}^{ - 20}}} \right)\left( {5 \times {{10}^4}} \right)} }}\]
Grouping the terms using parentheses, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {\left( {2 \times 91 \times 16 \times 5} \right)\left( {{{10}^{ - 32}} \times {{10}^{ - 20}} \times {{10}^4}} \right)} }}\]
Simplifying the expression using the rule of exponents \[{a^b} \times {a^c} = {a^{b + c}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {\left( {2 \times 91 \times 16 \times 5} \right)\left( {{{10}^{ - 32 - 20 + 4}}} \right)} }}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {\left( {2 \times 91 \times 16 \times 5} \right)\left( {{{10}^{ - 48}}} \right)} }}\end{array}\]
Multiplying the terms in the equation, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {14560\left( {{{10}^{ - 48}}} \right)} }}\]
The square root of a number can be written as \[\sqrt x = {x^{\dfrac{1}{2}}}\].
Rewriting the square root, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560 \times {{10}^{ - 48}}} \right)}^{\dfrac{1}{2}}}}}\]
If two or more numbers with different bases and same exponent are multiplied, the product can be written as \[{a^m} \times {b^m} = {\left( {ab} \right)^m}\]. The equation \[{\left( {ab} \right)^m} = {a^m} \times {b^m}\] is also true.
Using the rule of exponent \[{\left( {ab} \right)^m} = {a^m} \times {b^m}\] in the expression, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}} \times {{\left( {{{10}^{ - 48}}} \right)}^{\dfrac{1}{2}}}}}\]
If a number with an exponent is raised to another exponent, then the exponents are multiplied. This can be written as \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\].
Using the rule of exponent \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\] in the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}} \times \left( {{{10}^{ - 48 \times \dfrac{1}{2}}}} \right)}}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}} \times {{10}^{ - 24}}}}\end{array}\]
Rewriting the expression, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times \dfrac{{{{10}^{ - 37}}}}{{{{10}^{ - 24}}}}\]
Using the rule of exponents \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times {10^{ - 37 - \left( { - 24} \right)}}\]
Subtracting the terms in the exponent, we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times {10^{ - 37 + 24}}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times {10^{ - 13}}\end{array}\]
The square root of 14560 is approximately \[120.664825\].
Writing the square root of 14560, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx \dfrac{{6626}}{{120.664825}} \times {10^{ - 13}}\]
Dividing 6626 by \[120.664825\], we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 54.91244 \times {10^{ - 13}}\]
We will write the first number in the product as a number between 1 and 10.
Rewriting the number, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 5.491244 \times 10 \times {10^{ - 13}}\]
Using the rule of exponent \[{a^b} \times {a^c} = {a^{b + c}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 5.491244 \times {10^{ - 13 + 1}}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 5.5 \times {10^{ - 12}}\end{array}\]
\[\therefore \] We get the value of the expression \[\dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\] as approximately \[5.5 \times {10^{ - 12}}\].
Note: We rewrote \[54.91244\] as a number between 1 and 10 to keep the answer in the standard form.
A number is said to be in standard form if it is written as \[a \times {10^b}\], where \[a\] is a number between 1 and 10. The number \[5.5 \times {10^{ - 12}}\] is in standard form.
We have converted the decimal number in the fraction to get the whole number in both numerator and denominator. It is easier to apply rules of exponent to the whole number or fractions than applying it to decimal.
Formula Used:
We will use the following formulas:
1.\[{a^b} \times {a^c} = {a^{b + c}}\]
2.\[{a^m} \times {b^m} = {\left( {ab} \right)^m}\]
3.\[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\]
4.\[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\].
Complete step-by-step answer:
We will rewrite the expression \[\dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\] to find the required value.
Rewriting the decimal numbers as fractions, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{\dfrac{{6626}}{{1000}} \times {{10}^{ - 34}}}}{{\sqrt {2\left( {\dfrac{{91}}{{10}} \times {{10}^{ - 31}}} \right)\left( {\dfrac{{16}}{{10}} \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\]
Rewriting 50 as the product of 5 and 10, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{\dfrac{{6626}}{{1000}} \times {{10}^{ - 34}}}}{{\sqrt {2\left( {\dfrac{{91}}{{10}} \times {{10}^{ - 31}}} \right)\left( {\dfrac{{16}}{{10}} \times {{10}^{ - 19}}} \right)\left( {5 \times 10 \times {{10}^3}} \right)} }}\]
Rewriting 10 as 10 raised to the power 1, and 1000 as 10 raised to the power 3, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{\dfrac{{6626}}{{{{10}^3}}} \times {{10}^{ - 34}}}}{{\sqrt {2\left( {\dfrac{{91}}{{{{10}^1}}} \times {{10}^{ - 31}}} \right)\left( {\dfrac{{16}}{{{{10}^1}}} \times {{10}^{ - 19}}} \right)\left( {5 \times {{10}^1} \times {{10}^3}} \right)} }}\]
If two numbers with same base and different exponents are multiplied, the product can be written as \[{a^b} \times {a^c} = {a^{b + c}}\].
If two or more numbers with different bases and same exponent are divided, the quotient can be written as \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\].
Therefore, rewriting the expression using the rules of exponents \[{a^b} \times {a^c} = {a^{b + c}}\] and \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times \dfrac{{{{10}^{ - 34}}}}{{{{10}^3}}}}}{{\sqrt {2\left( {91 \times \dfrac{{{{10}^{ - 31}}}}{{{{10}^1}}}} \right)\left( {16 \times \dfrac{{{{10}^{ - 19}}}}{{{{10}^1}}}} \right)\left( {5 \times {{10}^{1 + 3}}} \right)} }}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 34 - 3}}}}{{\sqrt {2\left( {91 \times {{10}^{ - 31 - 1}}} \right)\left( {16 \times {{10}^{ - 19 - 1}}} \right)\left( {5 \times {{10}^{1 + 3}}} \right)} }}\end{array}\]
Adding the subtracting the terms in the exponents, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {2\left( {91 \times {{10}^{ - 32}}} \right)\left( {16 \times {{10}^{ - 20}}} \right)\left( {5 \times {{10}^4}} \right)} }}\]
Grouping the terms using parentheses, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {\left( {2 \times 91 \times 16 \times 5} \right)\left( {{{10}^{ - 32}} \times {{10}^{ - 20}} \times {{10}^4}} \right)} }}\]
Simplifying the expression using the rule of exponents \[{a^b} \times {a^c} = {a^{b + c}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {\left( {2 \times 91 \times 16 \times 5} \right)\left( {{{10}^{ - 32 - 20 + 4}}} \right)} }}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {\left( {2 \times 91 \times 16 \times 5} \right)\left( {{{10}^{ - 48}}} \right)} }}\end{array}\]
Multiplying the terms in the equation, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{\sqrt {14560\left( {{{10}^{ - 48}}} \right)} }}\]
The square root of a number can be written as \[\sqrt x = {x^{\dfrac{1}{2}}}\].
Rewriting the square root, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560 \times {{10}^{ - 48}}} \right)}^{\dfrac{1}{2}}}}}\]
If two or more numbers with different bases and same exponent are multiplied, the product can be written as \[{a^m} \times {b^m} = {\left( {ab} \right)^m}\]. The equation \[{\left( {ab} \right)^m} = {a^m} \times {b^m}\] is also true.
Using the rule of exponent \[{\left( {ab} \right)^m} = {a^m} \times {b^m}\] in the expression, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}} \times {{\left( {{{10}^{ - 48}}} \right)}^{\dfrac{1}{2}}}}}\]
If a number with an exponent is raised to another exponent, then the exponents are multiplied. This can be written as \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\].
Using the rule of exponent \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\] in the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}} \times \left( {{{10}^{ - 48 \times \dfrac{1}{2}}}} \right)}}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626 \times {{10}^{ - 37}}}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}} \times {{10}^{ - 24}}}}\end{array}\]
Rewriting the expression, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times \dfrac{{{{10}^{ - 37}}}}{{{{10}^{ - 24}}}}\]
Using the rule of exponents \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\], we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times {10^{ - 37 - \left( { - 24} \right)}}\]
Subtracting the terms in the exponent, we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times {10^{ - 37 + 24}}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} = \dfrac{{6626}}{{{{\left( {14560} \right)}^{\dfrac{1}{2}}}}} \times {10^{ - 13}}\end{array}\]
The square root of 14560 is approximately \[120.664825\].
Writing the square root of 14560, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx \dfrac{{6626}}{{120.664825}} \times {10^{ - 13}}\]
Dividing 6626 by \[120.664825\], we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 54.91244 \times {10^{ - 13}}\]
We will write the first number in the product as a number between 1 and 10.
Rewriting the number, we get
\[ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 5.491244 \times 10 \times {10^{ - 13}}\]
Using the rule of exponent \[{a^b} \times {a^c} = {a^{b + c}}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 5.491244 \times {10^{ - 13 + 1}}\\ \Rightarrow \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }} \approx 5.5 \times {10^{ - 12}}\end{array}\]
\[\therefore \] We get the value of the expression \[\dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2\left( {9.1 \times {{10}^{ - 31}}} \right)\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {50 \times {{10}^3}} \right)} }}\] as approximately \[5.5 \times {10^{ - 12}}\].
Note: We rewrote \[54.91244\] as a number between 1 and 10 to keep the answer in the standard form.
A number is said to be in standard form if it is written as \[a \times {10^b}\], where \[a\] is a number between 1 and 10. The number \[5.5 \times {10^{ - 12}}\] is in standard form.
We have converted the decimal number in the fraction to get the whole number in both numerator and denominator. It is easier to apply rules of exponent to the whole number or fractions than applying it to decimal.
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