Question 319: Dual Nature of Matter and Radiation

Question

When photons of energy hv fall on a photosensitive surface of work function $\mathrm{E}_0$, photoelectrons of maximum energy $k$ are emitted. If the frequency of radiation is doubled the maximum kinetic energy will be equal to ( $\mathrm{h}=$ Planck's constant)

Accepted Answer

The photoelectric effect can be described by the equation:
$$ K_{	ext{max}} = h
u - E_0 $$
where:
$ K_{	ext{max}} $ is the maximum kinetic energy of the emitted photoelectrons,
$ h $ is Planck's constant,
$ 
u $ is the frequency of the incident photons,
$ E_0 $ is the work function of the photosensitive surface.
Initially, photons of energy $ hv $ fall on the surface, yielding the maximum kinetic energy:
$$ k = h
u - E_0 $$
If the frequency of the radiation is doubled, the frequency becomes $ 2
u $. The new maximum kinetic energy $ K'_{	ext{max}} $ is:
$$ K'_{	ext{max}} = h(2
u) - E_0 = 2h
u - E_0 $$
Substitute the expression for $ k $ from the first equation:
$$ 2h
u - E_0 = (h
u - E_0) + h
u = k + h
u $$
Therefore, the maximum kinetic energy when the frequency is doubled would be:
Option D: $ k + h
u $

Answer

k

Answer

2 k

Answer

$\mathrm{k}+\mathrm{E}_0$

Answer

$\mathrm{k}+\mathrm{h} 
u$

Dual Nature of Matter and Radiation

When photons of energy hv fall on a photosensitive surface of work function \mathrm{E}_0, photoelectrons of maximum energy k are emitted. If the frequency of radiation is doubled the maximum kinetic energy will be equal to ( \mathrm{h}= Planck's constant)

Choose the correct answer: