Question

Consider 3^rd orbit of He⁺ (Helium), using non-relativistic approach, the speed of electron in this orbit will be $[givenK=9\times {10}^{9}cons\tan t,Z=2andh(Planck\text{'}scons\tan t)=6.6\times {10}^{-34}Js]$

• A. $1.46\times {10}^{6}m/s$
• B. $0.73\times {10}^{6}m/s$
• C. $3.0\times {10}^{8}m/s$
• D. $2.92\times {10}^{6}m/s$

Answer 1

The correct option is A

$1.46\times {10}^{6}m/s$

Step - 1: Given

$K=9\times {10}^{9}cons\tan t,Z=2andh(Planck\text{'}scons\tan t)=6.6\times {10}^{-34}Js$

$n=3,for{3}^{rd}orbit$ and $massofelectron{m}_e=9.1\times {10}^{-31}kg$

Step - 2: Formula

$E_n=-13.6\frac{Z^2}{n^2}eV$ and

$KE=\frac{1}{2}{m}_e{v}^2$

Step - 3: Solution

By substituting all values, we will get $$\begin{gathered} E_3=-13.6\times \frac{2^2}{3^2}eV \\ {E}_3=-13.6\times \frac{4}{9}\times 1.6\times {10}^{-19}J \\ {E}_3=9.7\times {10}^{-19}J \end{gathered}$$

and Kinetic energy of 3^rd orbit = E₃

$$\begin{gathered} \frac{1}{2}{m}_e{v}^2=E_3 \\ \frac{1}{2}\times 9.1\times {10}^{-31}\times v^2=9.7\times {10}^{-19} \\ v=\sqrt{\frac{2\times 9.7\times {10}^{-19}}{9.1\times {10}^{-31}}} \\ v=1.46\times {10}^{6}m/s \end{gathered}$$

Hence option A is correct.