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