Atomic Structure Question 386
Question: An electron, $ e_1 $ is moving in the fifth stationary state, and another electro $ e_2 $ is moving in the fourth stationary state. The radius of orbit of electron $ e_1 $ is five times the radius of orbit of electron, $ e_2 $ calculate the ratio of velocity of electron $ e_1(v_1) $ to the velocity of electron $ e_2(v_2) $
Options:
A) 5 : 1
B) 4 : 1
C) 1 : 5
D) 1 : 4
Show Answer
Answer:
Correct Answer: D
Solution:
- From the expression of Bohr’s theory, we know that $ m_{e}v_1r_1=n_1\frac{h}{2\pi } $
$ \And ,m_{e}v_2r_2=n_2\frac{h}{2\pi } $
$ \frac{m_{e}v_1r_1}{m_{e}v_2r_2}=\frac{n_1}{n_2}\frac{h}{2\pi }\times \frac{2\pi }{h} $ Given, $ r_1=5r_2,,n_1=5,n_2=4 $
$ \frac{m_{e}\times v_1\times 5r_2}{m_{e}\times v_2\times r_2}=\frac{5}{4} $
$ \Rightarrow \frac{v_1}{v_2}=\frac{5}{4\times 5}=\frac{1}{4}=1 : 4 $
