Q. In a stationary hydrogen atom, an electron jumps from n = 3 to n = 1. The recoil speed of the hydrogen atom is about

(a) 4 m/s

(b) 4 cm/s

(c)4 mm/s

(d) 4 × 10^{-4} m/s

Ans: (a)

Solution : $\displaystyle \frac{1}{\lambda} = R(\frac{1}{1^2} – \frac{1}{3^2}) $

$\displaystyle \frac{1}{\lambda} = R(1 – \frac{1}{9}) $

$\displaystyle \frac{1}{\lambda} =\frac{8R}{9} $

$\displaystyle \lambda =\frac{9}{8R} $ …(i)

Momentum , P = mv

$\displaystyle \frac{h}{\lambda} = m v $

$\displaystyle v = \frac{h}{\lambda m} $

$\displaystyle v = \frac{h}{\frac{9}{8R} m} $ ;

( on putting the value of λ from(i))

v = 8Rh/9m

By putting the Standard values of R , h & m we get,

v = 4 m/s