Ncert Exercises & Solutions

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

The energy of the incident photon (E) is given by,

$E = \frac{hc}{λ}$
$= \frac{(6.626 x 10^{- 34}) (3.0 x 10^{8} {ms}^{- 1})}{(150 x 10^{- 12} m)}$
$= 1.3252 x 10^{- 15} J$
$= 13.252 x 10^{- 16} J$

The energy of the electron ejected (K.E)

$= \frac{1}{2} {mv}^{2}$
$= \frac{1}{2} (9.10939 x 10^{- 31} kg) (1.5 x 10^{7} {ms}^{- 1})$
= 10.2480 × 10–17 J
= 1.025 × 10–16 J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

= E – K.E
= 13.252 × 10–16 J – 1.025 × 10–16 J
= 12.227 × 10–16 J

$= \frac{12.227 x 10^{- 16}}{1.602 x 10^{- 19}} eV$
$= 7.6 x 10^{3} eV$
$\frac{5 λ_{0} - 2000}{4 λ_{0} - 2000} = {(\frac{5.35}{2.55})}^{2} = \frac{28.6225}{6.5025}$
$\frac{5 λ_{0} - 2000}{4 λ_{0} - 2000} = 4.40177$
$17.6070 λ_{0} - 5 λ_{0} = 8803.537 - 2000$
$λ_{0} = \frac{6805.537}{12.607}$
$λ_{0} = 539.8 nm$
$λ_{0} = 540 nm$

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