NCERT Class XI Chemistry Structure of Atom Solutions

Question : 54

Total: 67

If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107ms–1, calculate the energy with which it is bound to the nucleus.

Solution:

Photon of wavelength = 150 pm = 150 × 10–12m
Energy of photon (E) =

6.626×10−34×3×108

150×1012

= 0.1325 × 10−14
= 13.25 × 10−16J
Energy of the ejected electron =

mv2 =

× 9.11 × 10−31×(1.5×107)2
Energy with which the electron is bound to the nucleus
= (13.25 × 10–16 – 1.025 × 10–16) J = 12.225 × 10–16J
=

12.225×10−16

1.602×10−19

= 7.63 × 103 eV
[Since 1.602 × 10−19J = 1 eV]

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