Question

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is

Answer 1

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Answer 》 Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is 1.6 × 10^5 Hz

Full solution 》》》》》》

Energy realised from emission of electron

E= (-3.4) - (-13.6)= 10.2 ev

From Photo electric equation work function

ø = E- ev = hv

v = E - ev / h

(10.2) - (3.57)e / 6.67 × 10 ^34

v = 6.63 × 1.6 × 10 ^19 / 6.67 × 10 ^34

= 1.6 × 10^5 Hz answer

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Answer 2

Answer:

For hydrogen atom the energy in nth state is

En=

$\frac{ - 1.36ev}{ {n}^{2} }$

First excited state means second normal state i.e

$n = 2$

ground state is n=1 so the energy of emitted photon will be

E= -(E1-E2)

−13.6eV(−1+1/4)=10.2 ev

Stopping potential

$v = 3.75volt$

on using Einstein equation, eV =E−W, W is work function

so W=(10.2−3.57)eV=6.63eV=6.63×1.6×10^-19joule

=1.0608×10^-18 Joule

the threshold frequency , f=W/h......(1)

h is Planck's constant h=6.62×10^-34Js

putting these values in equation-1 , we get

f=1.6×10^15Hz