An electron jumps from the first excited state to the ground stage of hydrogen atom..What will be the percentage change in the speed of electron ?

Options

A. 0.25
B. 0.5
C. 1
D. 2

Answer

B

Explanation

As `v_(n)prop(1)/(n)`
so, `(v_(2))/(v_(1))=(1)/(2)` i.le., `v_(2)=(v_(1))/(2)`
`Deltav=v_(1)-v_(2)=(v_(1))/(2)`
Hence, % change=50%