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The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from ′λ′1 to ′λ′2. Derive the expressions for the threshold wavelength ′λ′0 and work function for the metal surface

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Solution

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hν=W0+KE

where ν is the frequency of incident light, W0 is the work potential of the metal, and KE is the maximum kinetc energy of released electron.

The above equations explains the following results:

1. If ν<νo, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if ν>νo.

2. The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectron depends only on the frequency of incident light.

1. If ν<νo, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if ν>νo.

2. The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectron depends only on the frequency of incident light.

From given data,

hcλ1=KE+Wo

hcλ2=2KE+Wo

Eliminating KE, work function is given by:

Wo=2hcλ1−hcλ2

Threshold wavelength is found by:

hcλo=Wo

Using above equations,

λo=λ1λ22λ2−λ1

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