When photons of energy hv fall on a photosensitive surface of work function E 0 , photoelectrons of maximum energy k are emitted. If the frequency of radiation is doubled the maximum kinetic energy

Question

When photons of energy hv fall on a photosensitive surface of work function E 0 ​ , photoelectrons of maximum energy k are emitted. If the frequency of radiation is doubled the maximum kinetic energy will be equal to ( h = Planck's constant)

TestHub · Accepted Answer

Using Einstein's photoelectric equation for the initial case:

$k = h ν - E_{0}$

When the frequency is doubled, the new energy of the incident photons is $2 h ν$ . The new maximum kinetic energy $k^{'}$ is given by:

$k^{'} = 2 h ν - E_{0}$

This can be rewritten as:

$k^{'} = h ν + (h ν - E_{0})$

Substituting $h ν - E_{0} = k$ into the equation:

$k^{'} = h ν + k$

Answer: $k + h v$

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