Question

The wavelength corresponding to a transition when electron falls from certain quantum level to the ground state of an $H{e}^{+}$ ion is 24.31 nm. Find the ratio of velocity of the electron in the next quantum level to that of velocity of light?

• A. 1:228
• B. 1:524
• C. 1:344
• D. 1: 160

Answer 1

The correct option is C 1:344

Given: $\lambda =24.31nm$
we know,
$\frac{1}{\lambda }=R_H\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]\frac{1}{24.31\times {10}^{-7}}=109678\times 2^2(\frac{1}{1^2}-\frac{1}{n_2^2})$
$n^2\approx 16.05$
$n\approx 4$
$n_2=4$ as n has to be an integer
so, next level $n=5$
velocity of electron in $5^{th}$ level of $H{e}^{+},V_n=2.18\times {10}^8\frac{Z}{n}cm/s=2.18\times {10}^8\times \frac{2}{5}$ = $8.72\times {10}^{7}cm/s$
so ratio will be:
velocity of the electron in $5^{th}$level : velocity of light
= $8.72\times {10}^7:3\times {10}^{10}$
= $1:344$