MCQ
A hydrogen atom emits a photon corresponding to an electron transition from $n = 5$ to $n = 1$. The recoil speed of hydrogen atom is almost......$ms^{-1}$ (mass of proton $\approx 1.6 \times 10^{-27}kg$).
  • A
    $10$
  • B
    $2 \times 10^{-2} $
  • $4$
  • D
    $8 \times 10^2 $

Answer

Correct option: C.
$4$
c
(c) The Hydrogen atom before the transition was at rest. Therefore from conservation of momentum.
${p_{H - atom}} = {p_{photon}} = \frac{{{E_{radiated}}}}{c} = \frac{{13.6\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)eV}}{c}$
$1.6 \times {10^{ - 27}} \times v = \frac{{13.6\left( {\frac{1}{{{1^2}}} - \frac{1}{{{5^2}}}} \right) \times 1.6 \times {{10}^{ - 19}}}}{{3 \times {{10}^8}}}$
==> $v = 4.352\, m/s \approx  4\, m/sec.$

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