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Question

Light of wavelengthλ1=340nm and λ2=540nm are incident on a metallic surface. If the ratio of the speed of the electrons ejected is 2. The work function of the metal is.


A

1eV

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B

1.85eV

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C

1.5eV

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D

2eV

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Solution

The correct option is B

1.85eV


Step 1. Given data:

  1. The wavelength of incident lightλ1=340nm=340×10-9m
  2. The wavelength of another incident lightλ2=540nm=540×10-9m
  3. If the ratio of the speed of the electrons ejectedv1v2=2
  4. Planck's constanth=6·6×10-34m2kgs-1
  5. Speed of lightc=3×108ms-1

Step 2. Formula used:

Energyofphoton=workfunction+maximumenergyofemittedelectron

hcλ=ϕ+12mv2

Where, h is planck's constant

Step 3. Calculating the work function of metal:

The work function is the energy required to eject an electron from a metal surface, which can be calculated as follows,

hcλ1=ϕ+12mv21and,hcλ2=ϕ+12mv22asv1=2v2

Then from the above equations, we can write,

hcλ1-ϕhcλ2-ϕ=v1v22=4

Now, by equating,

hcλ1-ϕ=4hcλ2-ϕ3ϕ=4hcλ2-hcλ13ϕ=hc4λ2-1λ1=6·6×10-34×3×108×4540-1340×109ϕ=5·6×10-19J(as,1eV=1·6×10-19J)=1·842eV1·85eV

Thus, the work function of the metal is1·85eV.

Hence, option B is the correct answer.


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