KVPY SX SB Exam 2013 Question Paper

Question : 94 of 120

Marks: +1, -0

A singly ionized helium atom in an excited state (n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is -13.6 eV, the energy

(E_f)

and quantum number (n) of the resulting state respectively.

$E_f=-13.6\eV, n=1$
$E_f= -6.0 \eV, n=3$
$E_f =-6.0\eV,n=2$
$E_f= -13.6\eV, n=2$

Validate

Solution:

2.6=13.6z^2[1/n^2-1/4^2]

2.6/{13.6×4}=2^2[1/n^2-1/4^2]

2.6/{13.6×4}=1/n^2-1/16

n = 3 now Energy

E={13.6Z^2}/n^2\eV

=-{13.6×4}/9\eV

= -6eV

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