An electron of kinetic energy $E_{0}$ is scattered by an atomic hydrogen sample in ground state. The minimum value of $E_{0}$ , so that a photon of wavelength $656.3 n m$ may be emitted by $H -$ atom is :

12.09 e V

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13.6 e V

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14.6 e V

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Solution

The correct option is A $12.09 e V$
Given : $Z = 30$ $n = 1$
Energy required $E = \frac{13.6 Z^{2}}{n^{2}}$
$∴$ $E = \frac{13.6 \times 30^{2}}{1} = 12.24$ keV

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An electron of kinetic energy E0 is scattered by an atomic hydrogen sample in ground state. The minimum value of E0, so that a photon of wavelength 656.3 nm may be emitted by H−atom is :

The correct option is A 12.09 eVGiven : Z=30 n=1Energy required E=13.6Z2n2∴ E=13.6×3021=12.24 keV

An electron of kinetic energy $E_{0}$ is scattered by an atomic hydrogen sample in ground state. The minimum value of $E_{0}$ , so that a photon of wavelength $656.3 n m$ may be emitted by $H -$ atom is :

The correct option is A $12.09 e V$
Given : $Z = 30$ $n = 1$
Energy required $E = \frac{13.6 Z^{2}}{n^{2}}$
$∴$ $E = \frac{13.6 \times 30^{2}}{1} = 12.24$ keV