Question: Q. 12. A monochromatic light source of power 5 mW emits 8×1015 photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this set up is 2 V. Calculate the work function of the metal.

A [Delhi I, II, III 2016]

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Solution:

Ans.

P=5×103 W n=PE

E=Pn=6.25×1019 J1/2

E=3.9eV1/2 W0=EeV01/2

=(3.92)eV

W0=1.9eV

Detailed Answer :

[CBSE Marking Scheme 2016]

Power of monochromatic light source

= total energy released per second

= number of photons emits per second

× energy of one photon

Number of photons emits per second =8×1015

Power of monochromatic light source =5 mW

Hence, energy of one photon,

hv=5×1038×1015 =0.625×1018 J =0.625×10181.6×1019=3.9eV

K.Emax =eV0; where V0 is stopping potential.

=V0eV=2eV(V0=2 Volts ) K.E.=hvϕ0 2=3.9ϕ2 ϕ0=(3.92)eV ϕ0=1.9eV

Answering Tips

  • First calculate the energy of one photon and convert it to eV.

Short Answer Type Questions-II



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