KVPY SX SB Exam 20141 Question Paper

Question : 99 of 120

Marks: +1, -0

When ultraviolet radiation of a certain frequency falls on a potassium target, the photoelectrons released can be stopped completely by a retarding potential of 0.6 V. If the frequency of the radiation is increased by 10%, this stopping potential rises to 0.9 V. The work function of potassium is

2.0 eV
2.4 eV
3.0 eV
2.8 eV

Validate

Solution:

KE_\max= e × V_\etarding

= e × 0.6
= 0.6 eV
Photon energy = hf = E
When frequency increase by 10% energy of photon also increases by 10%
New energy = E' = 1.1 E
New

KE_\max. = e × V_\retarding = e × 0.9 = 0.9

eV
Einstein photoelectric equation

hf = KE_\max + ϕ

E = 0.6 + ϕ

…(1)

1.1 E = 0.9 + ϕ

…(2)

1.1 = {0.9+ϕ}/{0.6+ϕ}

1.1 ϕ + 0.66 = 0.9 + ϕ

0.1 ϕ = 0.24

ϕ = 2.4

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