Question

Let a light wavelength and intensity I strike a metal surface to emit x electrons per second. Average energy of each electron is the unit. What will happen to x and y when
(a)$\lambda $ is halved
(b) intensity I is doubled

Answer 1

Hint:We will use the photoelectric effect to determine the effect on x and y. According to the photoelectric effect, the light falling on the metal surface causes the ejection from the surface. The number of emitted electrons depends upon the frequency. The kinetic energy depends upon the frequency.
Formula used: ${{h\nu }}\,{{ = }}\,{{h}}{{{\nu }}_{{o}}}\,{{ + }}\,\dfrac{{{1}}}{{{2}}}{{m}}{{{v}}^{{2}}}$

Complete answer :
When light strikes the surface of the metal and an electron ejection takes place is known as the photoelectric effect. The ejected electron is known as photoelectrons.

The process takes place in two parts:
Some energy of emitted light causes the ionization of the metal so, electrons get free form the metal and some part of the energy provides the kinetic energy to the electron so, electrons lose the surface of the metal. If the energy of the emitted photon is very less than it can free the electron from the metal but not from the surface.

${{h\nu }}\,{{ = }}\,{{h}}{{{\nu }}_{{o}}}\,{{ + }}\,\dfrac{{{1}}}{{{2}}}{{m}}{{{v}}^{{2}}}$

${{h\nu }}$is the absolute energy of electrons.
${{h}}{{{\nu }}_{{o}}}$ is the threshold energy.
$\dfrac{{{1}}}{{{2}}}{{m}}{{{v}}^{{2}}}$is the average energy of ejected electrons.

Average energy of each electron is y so,
$\Rightarrow \,{{y}}\,{{ = }}\,\,{{h\nu }}\,{{ - h}}{{{\nu }}_{{o}}}$
$\Rightarrow {{\nu }}\,{{ = }}\,\dfrac{{{c}}}{{{\lambda }}}$
$\Rightarrow\,{{y}}\,{{ = }}\,\,\dfrac{{{{hc}}}}{{{\lambda }}}\,{{ - h}}{{{\nu }}_{{o}}}$

(a)$\lambda $ is halved
X defines the electrons emitted per second which is the rate. Change in wavelength does not affect the rate of electron ejection. As the average energy of ejected electron y is inversely proportional to the wavelength so, when $\lambda $ is halved the average energy increases. So, When$\lambda $ is halved x does not get affected and y increases.

(b) Intensity I is doubled
X defines the electrons emitted per second which is the rate. Intensity defines as the power per unit area. So, increasing the intensity, the electron ejected per unit area will increase.
On increasing intensity only rate increases, not the energy so, the average energy of the ejected electron will not be affected.

Note:${{h}}{{{\nu }}_{{o}}}$is the threshold energy. It is also known as work function. It is the energy required to free the electron from the metal. The energy required to free the electron from the metal is known as binding energy. When the wavelength is halved the average energy increases but we cannot say that average energy will be double because average energy also depends upon the work function.