SATHEE: Atoms And Nuclei Question 328

Atoms And Nuclei Question 328

Question: If radiation corresponding to first line of “Balmer series” of $H e^{+}$ ion knocked out electron from 1st excited state of H atom, the kinetic energy of ejected electron from H atom would be (eV) - [Given $E_{n} = - \frac{Z^{2}}{n^{2}} (13.6 e V)$ ]

Options:

A) 4.155 eV

B) 8.310 eV

C) 2.515 eV

D) 5.550 eV

Show Answer

Answer:

Correct Answer: A

Solution:

Energy of photon corresponding to first line of Balmer series $= (13.6) {(2)}^{2} [\frac{1}{4} - \frac{1}{9}]$ Energy need to eject electron from n=2 level in H atom $= (13.6) (\frac{1}{4})$ So, required kinetic energy $= (13.6) [(\frac{1}{4} - \frac{1}{9}) - (\frac{1}{4})] e V$ $= 13.6 \times (\frac{11}{36}) = 4.155 e V$

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Atoms And Nuclei Question 328

Question: If radiation corresponding to first line of “Balmer series” of He+ ion knocked out electron from 1st excited state of H atom, the kinetic energy of ejected electron from H atom would be (eV) - [Given En=−Z2n2(13.6eV) ]

Options:

Answer:

Solution:

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Question: If radiation corresponding to first line of “Balmer series” of $H e^{+}$ ion knocked out electron from 1st excited state of H atom, the kinetic energy of ejected electron from H atom would be (eV) - [Given $E_{n} = - \frac{Z^{2}}{n^{2}} (13.6 e V)$ ]