Question

Uranium has two isotopes of masses $235$ and $238$ units. If both of them are present in uranium hexafluoride gas, find the percentage ratio of difference in rms velocities of two isotopes to the rms velocity of heavier isotopes?
A. $1.64$
B. $0.064$
C. $0.64$
D. $6.4$

Answer 1

Hint: The masses of the two isotopes of the uranium metal is given. The velocity of the root mean square is given by an equation which states that velocity of rms is inversely proportional to the mass of the isotope. By taking a relation between the velocity rms of the two masses. The relation between the velocity rms of the two masses is derived. By using this the ratio of difference between the rms velocities of two isotopes to the rms velocity of heavier isotopes is calculated.

Useful formula:
The velocity of root mean square,
${V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
Where, ${V_{rms}}$ is the rms velocity, $R$ is the gas constant, $T$ is absolute temperature in kelvin and $M$ is the molar mass of the isotope.

Given data:
The mass of the first isotope, ${M_1} = 235\;units$
The mass of the second isotope, ${M_2} = 238\;units$

Complete step by step solution:
Assume that,
The rms velocity of first isotope is ${V_{1\left( {rms} \right)}}$
The rms velocity of second isotope is ${V_{2\left( {rms} \right)}}$

The rms velocity of first isotope is given by,
${V_{1\left( {rms} \right)}} = \sqrt {\dfrac{{3RT}}{{{M_1}}}} $
Here, the rms velocity is inversely proportional to the square root of mass,
Thus,
${V_{1\left( {rms} \right)}} \propto \dfrac{1}{{\sqrt {{M_1}} }}\;...........................................\left( 1 \right)$

The rms velocity of second isotope is given by,
${V_{2\left( {rms} \right)}} = \sqrt {\dfrac{{3RT}}{{{M_2}}}} \;$
Here, the rms velocity is inversely proportional to the square root of mass,
Thus,
${V_{2\left( {rms} \right)}} \propto \dfrac{1}{{\sqrt {{M_2}} }}\;...........................................\left( 2 \right)$
From the equation (1) and (2), we get
$
  \dfrac{{{V_{1\left( {rms} \right)}}}}{{{V_{2\left( {rms} \right)}}}} = \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }} \\
  \dfrac{{{V_{1\left( {rms} \right)}}}}{{{V_{2\left( {rms} \right)}}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \\
 $
By substituting the values of ${M_1}$ and ${M_2}$ in above relation, we get
$
  \dfrac{{{V_{1\left( {rms} \right)}}}}{{{V_{2\left( {rms} \right)}}}} = \sqrt {\dfrac{{238\;units}}{{235\;units}}} \\
  \dfrac{{{V_{1\left( {rms} \right)}}}}{{{V_{2\left( {rms} \right)}}}} = \sqrt {1.01276} \\
 $
Taking the root to the value inside the square root,
$
  \dfrac{{{V_{1\left( {rms} \right)}}}}{{{V_{2\left( {rms} \right)}}}} = 1.0064 \\
  {V_{1\left( {rms} \right)}} = 1.0064 \times {V_{2\left( {rms} \right)}}\;............................\left( 3 \right) \\
 $

The ratio between the difference in rms velocities to rms velocity of heavier isotope is $\dfrac{{\left( {{V_{1\left( {rms} \right)}} - {V_{2\left( {rms} \right)}}} \right)}}{{{V_{2\left( {rms} \right)}}}}$.
By substituting the equation (3) in above relation, we get
$\dfrac{{\left( {{V_{1\left( {rms} \right)}} - {V_{2\left( {rms} \right)}}} \right)}}{{{V_{2\left( {rms} \right)}}}} = \dfrac{{\left( {1.0064 \times {V_{2\left( {rms} \right)}}} \right) - {V_{2\left( {rms} \right)}}}}{{{V_{2\left( {rms} \right)}}}}$
Taking out the common term outside the bracket in RHS,
$\dfrac{{\left( {{V_{1\left( {rms} \right)}} - {V_{2\left( {rms} \right)}}} \right)}}{{{V_{2\left( {rms} \right)}}}} = \dfrac{{{V_{2\left( {rms} \right)}} \times \left( {1.0064 - 1} \right)}}{{{V_{2\left( {rms} \right)}}}}$
By cancelling the term ${V_{2\left( {rms} \right)}}$ in RHS, we get
$
  \dfrac{{\left( {{V_{1\left( {rms} \right)}} - {V_{2\left( {rms} \right)}}} \right)}}{{{V_{2\left( {rms} \right)}}}} = \left( {1.0064 - 1} \right) \\
  \dfrac{{\left( {{V_{1\left( {rms} \right)}} - {V_{2\left( {rms} \right)}}} \right)}}{{{V_{2\left( {rms} \right)}}}} = 0.0064 \\
 $
The percentage of the ratio between the difference in rms velocities to rms velocity of heavier isotope is given by,
$\dfrac{{\left( {{V_{1\left( {rms} \right)}} - {V_{2\left( {rms} \right)}}} \right)}}{{{V_{2\left( {rms} \right)}}}} \times 100 = 0.0064 \times 100 = 0.64$

Hence, the option (C) is correct.

Note: From this it is clear that there is a decrease in the rms velocity of the isotope when the mass of the isotope gets increased. The value of gas constant and the absolute temperature are constant for both of the isotopes in finding the rms velocity of each. Since, both the isotopes are present in hexafluoride gas, it is required to find the relation between the two isotopes.