Answer
Verified
91.5k+ views
Hint: Recall how the velocity of electrons in the orbits of the Bohr’s model is related to the atomic number of the element and the quantum number of the orbital that specific electron is located in.
Complete step by step solution:
We know that the velocity of an electron as defined by Bohr’s model is the product of the velocity of an electron in the first orbit of a hydrogen atom and the atomic number of the element, this value is divided by the quantum number of the orbital the electron is present in. This formula is denoted as:
\[{{v}_{e}}=2.19\times {{10}^{6}}m{{s}^{-1}}\times \dfrac{z}{n}\]
Where, ${{v}_{e}}$ denotes the velocity of the electron, $z$ denotes the atomic number, and $n$ denotes the quantum number of the orbital. The constant given is the velocity of an electron in the first orbital of the hydrogen atom. In this question, we have to find $n$.
The ratio of the velocity of light to the velocity of the electron is given to us as 275:1. So, it will be shown as:
\[\dfrac{{{v}_{e}}}{c}=\dfrac{1}{275}\]
Where, $c$is the speed of light i.e. $3\times {{10}^{8}}m/s$. Rearranging the above equation, we get:
\[{{v}_{e}}=\dfrac{c}{275}\]
We will substitute this value, along with the value of $z$ as 1, since the atomic number of hydrogen is 1, in the formula for the velocity of an electron. Then we will rearrange the formula and solve for $n$. We will solve the equation as follows:
\[\begin{align}
& \dfrac{3\times {{10}^{8}}}{275}m{{s}^{-1}}=2.19\times {{10}^{19}}m{{s}^{-1}}\times \dfrac{1}{n} \\
& n=\dfrac{2.19\times {{10}^{6}}\times 275}{3\times {{10}^{8}}} \\
& n=200.75\times {{10}^{-2}} \\
& n=2 \\
\end{align}\]
Hence, the answer to this question is B ‘2’.
Additional information:
The value $2.18\times {{10}^{6}}$ is defined as the velocity of the electron if it was present in the first orbital of the hydrogen atom. We know that an electron shows both waves as well as the particle nature, so we can conclude that the circumference of the circular orbit of the electron should be an integral multiple of its wavelength. So, we get the formula as:
\[2\pi r=n\lambda \]
Where, all the variables take their standard meanings. We know the de Broglie hypothesis defines the de Broglie wavelength as:
\[\lambda =\dfrac{h}{mv}\]
Here, $v$ is the velocity and the other variables take their standard meanings. We will substitute the value for lambda in the earlier equation and rearrange the equation to solve for $v$. So, the equation that we obtain is:
\[\begin{align}
& 2\pi r=n\times \dfrac{h}{mv} \\
& v=\dfrac{nh}{2\pi rm} \\
\end{align}\]
Now, we will substitute the known values of all the variables and obtain the value for velocity.
\[\begin{align}
& n=1 \\
& h=6.63\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}} \\
& \pi =3.142 \\
& r=0.529\times {{10}^{-10}}m \\
& m=9.1\times {{10}^{-31}}kg \\
\end{align}\]
Now, we will substitute these values in the equation and obtain the value for velocity
\[\begin{align}
& v=\dfrac{1\times 6.63\times {{10}^{-34}}}{2\times 3.142\times 0.529\times {{10}^{-10}}\times 9.1\times {{10}^{-31}}} \\
& v=2.18\times {{10}^{6}}m/s \\
\end{align}\]
Note: While the ratio of the velocities is defined, we will not consider the units but as soon as the value of ${{v}_{e}}$ is determined in terms of the speed of light we have to consider the units of the velocity as $m/s$. Remember that the velocity is directly proportional to the atomic number and inversely proportional to how far away the electron I from the nucleus i.e. in which orbital it is present. You can find the velocity by proportionality too if you know the velocity of the electron in the first orbit of hydrogen.
Complete step by step solution:
We know that the velocity of an electron as defined by Bohr’s model is the product of the velocity of an electron in the first orbit of a hydrogen atom and the atomic number of the element, this value is divided by the quantum number of the orbital the electron is present in. This formula is denoted as:
\[{{v}_{e}}=2.19\times {{10}^{6}}m{{s}^{-1}}\times \dfrac{z}{n}\]
Where, ${{v}_{e}}$ denotes the velocity of the electron, $z$ denotes the atomic number, and $n$ denotes the quantum number of the orbital. The constant given is the velocity of an electron in the first orbital of the hydrogen atom. In this question, we have to find $n$.
The ratio of the velocity of light to the velocity of the electron is given to us as 275:1. So, it will be shown as:
\[\dfrac{{{v}_{e}}}{c}=\dfrac{1}{275}\]
Where, $c$is the speed of light i.e. $3\times {{10}^{8}}m/s$. Rearranging the above equation, we get:
\[{{v}_{e}}=\dfrac{c}{275}\]
We will substitute this value, along with the value of $z$ as 1, since the atomic number of hydrogen is 1, in the formula for the velocity of an electron. Then we will rearrange the formula and solve for $n$. We will solve the equation as follows:
\[\begin{align}
& \dfrac{3\times {{10}^{8}}}{275}m{{s}^{-1}}=2.19\times {{10}^{19}}m{{s}^{-1}}\times \dfrac{1}{n} \\
& n=\dfrac{2.19\times {{10}^{6}}\times 275}{3\times {{10}^{8}}} \\
& n=200.75\times {{10}^{-2}} \\
& n=2 \\
\end{align}\]
Hence, the answer to this question is B ‘2’.
Additional information:
The value $2.18\times {{10}^{6}}$ is defined as the velocity of the electron if it was present in the first orbital of the hydrogen atom. We know that an electron shows both waves as well as the particle nature, so we can conclude that the circumference of the circular orbit of the electron should be an integral multiple of its wavelength. So, we get the formula as:
\[2\pi r=n\lambda \]
Where, all the variables take their standard meanings. We know the de Broglie hypothesis defines the de Broglie wavelength as:
\[\lambda =\dfrac{h}{mv}\]
Here, $v$ is the velocity and the other variables take their standard meanings. We will substitute the value for lambda in the earlier equation and rearrange the equation to solve for $v$. So, the equation that we obtain is:
\[\begin{align}
& 2\pi r=n\times \dfrac{h}{mv} \\
& v=\dfrac{nh}{2\pi rm} \\
\end{align}\]
Now, we will substitute the known values of all the variables and obtain the value for velocity.
\[\begin{align}
& n=1 \\
& h=6.63\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}} \\
& \pi =3.142 \\
& r=0.529\times {{10}^{-10}}m \\
& m=9.1\times {{10}^{-31}}kg \\
\end{align}\]
Now, we will substitute these values in the equation and obtain the value for velocity
\[\begin{align}
& v=\dfrac{1\times 6.63\times {{10}^{-34}}}{2\times 3.142\times 0.529\times {{10}^{-10}}\times 9.1\times {{10}^{-31}}} \\
& v=2.18\times {{10}^{6}}m/s \\
\end{align}\]
Note: While the ratio of the velocities is defined, we will not consider the units but as soon as the value of ${{v}_{e}}$ is determined in terms of the speed of light we have to consider the units of the velocity as $m/s$. Remember that the velocity is directly proportional to the atomic number and inversely proportional to how far away the electron I from the nucleus i.e. in which orbital it is present. You can find the velocity by proportionality too if you know the velocity of the electron in the first orbit of hydrogen.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
A soldier with a machine gun falling from an airplane class 11 physics JEE_MAIN
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main