Question

The kinetic energy of an electron is $E$ when the incident wavelength is $\lambda$ . To increase the kinetic energy of the electron to $2E$ , the incident wavelength must be:
A. $2\lambda$
B. $\dfrac{\lambda }{2}$
C. $\dfrac{{hc\lambda }}{{\left( {E\lambda + hc} \right)}}$
D. $\dfrac{{hc\lambda }}{{\left( {2E\lambda + hc} \right)}}$

Answer 1

the photoelectric effect is the ejection of electrons from the surface of metal when light falls on the metal. Here, we will use the formula of Einstein’s equation for photoelectric effect to calculate the incident wavelength. Here, the kinetic energy of electrons $E$ will be increased to $2E$

<>bFORMULA USED:The Einstein’s equation for photoelectric effect is given by

$K.E. = \dfrac{{hc}}{\lambda } - W$

Here, $K.E.$ is the kinetic energy of electron, $h$ is the Planck’s constant, $c$ is the speed of light,
$\lambda$ is the wavelength of incident light and $W$ is the work function of the metal.

COMPLETE STEP BY STEP ANSWER:
Here, in the above question, when the light is incident with wavelength $\lambda$ the kinetic energy of an electron is $E$ . This is the concept of photoelectric effect.
The photoelectric effect is the phenomenon of the emission of electrons after the incident light strikes the metal.

Now, to calculate the incident wavelength, we will use the Einstein’s equation for photoelectric effect which is shown below

$K.E. = \dfrac{{hc}}{\lambda } - W$
Now, in the above equation, the kinetic energy of the electron is $E$ . Therefore, the above
equation will become

$E = \dfrac{{hc}}{\lambda } - W$
$\Rightarrow \,W = \dfrac{{hc}}{\lambda } - E$

Now, when the kinetic energy is increased to $2E$ . Therefore, the equation will become
$2E = \dfrac{{hc}}{{\lambda '}} - W$

Now, putting the value of $W$ in the above equation, we get
$2E = \dfrac{{hc}}{{\lambda '}} - \dfrac{{hc}}{\lambda } + E$
$\Rightarrow \,2E - E + \dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{\lambda '}}$
$\Rightarrow \,E + \dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{\lambda '}}$
$\Rightarrow \,\dfrac{{hc}}{{\lambda '}} = \dfrac{{E\lambda + hc}}{\lambda }$
$\Rightarrow \,\dfrac{{\lambda '}}{{hc}} = \dfrac{\lambda }{{E\lambda + hc}}$
$\Rightarrow \,\lambda ' = \dfrac{{hc\lambda }}{{E\lambda + hc}}$
Therefore, the kinetic energy is increased to $2E$ , the incident wavelength must be
$\dfrac{{hc\lambda }}{{E\lambda + hc}}$

Hence, option (C) is the correct option.

NOTE: Here, $\lambda$ is the wavelength of the electron when the kinetic energy of the electron is $E$ . On the other hand, $\lambda '$ is the wavelength of the electron when the kinetic energy of the electron is increased to $2E$ . Hence, we have calculated $\lambda '$ that is the incident radiation.