MCQ
Energy of an electron is given by $E =- 2.178 \times 10^{-18}\,J \, \left( {\frac{{{Z^2}}}{{{n^2}}}} \right) .$ Wavelength of light required to excite an electron in an hydrogen atom from level $n = 1$ to $n = 2$ will be :

$(h = 6.62 \times 10^{-34} \,Js$ and $c = 3.0 \times 10^8 \,ms^{-1})$

  • $1.214 \times 10^{-7} \,m$
  • B
    $2.816 \times 10^{-7}\, m$
  • C
    $6.500 \times 10^{-7}\, m$
  • D
    $8.500  \times 10^{-7}\, m$

Answer

Correct option: A.
$1.214 \times 10^{-7} \,m$
a
Let $\lambda$ is the wavelength of that photon then

$h c / \lambda=E_{2}-E_{1}$

and using value of given E, $E=\frac{h c}{\lambda}=2.178 \times 10^{-18} \times z^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right](z=1)$

$\therefore \lambda=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.178 \times 10^{-18}} \times \frac{4}{3}=1.214 \times 10^{-7} \mathrm{m}$

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