Question

When light of wavelength $\lambda $ is incident on a metal surface, the stopping potential is found to be ${V_0}$. When light of wavelength $n\lambda $ is incident on the same metal surface, the stopping potential is found to be $\dfrac{{{V_0}}}{{n + 1}}$. The threshold wavelength of the metal is ?

Answer 1

Hint: Using the values of the wavelengths and stopping potentials given in the problem apply the formula for work function. Next, keeping this formula as a base, construct two corresponding equations given by the photoelectric light equation using the given data for the incident lights with differing wavelengths and stopping potentials. Equate them and rearrange the terms using cross multiplication and obtain the value of threshold wavelength.

Complete step by step answer:
The wavelengths of the incident light and their corresponding stopping potentials are given in the above question. Let the threshold wavelength to be calculated be denoted as ${\lambda _0}$. This threshold wavelength is the maximum value of wavelength needed for the photoelectric effect to occur. Now, we need to formulate an equation for the work function of a metal. The work function of a metal is the minimum amount of energy required by an electron to just escape from its surface.
The formula applied is:
$\phi = \dfrac{{h \times c}}{{{\lambda _0}}}$
where, $\phi $ denotes the work function of a metal, $h$ denotes the Plank’s constant, $c$ denotes a constant that is the speed of light and ${\lambda _0}$ denotes the threshold wavelength.

Since $h$ and $c$ are constants they have constant values that have been already calculated. These values are:
$h = 6.626 \times {10^{ - 34}}$
$\Rightarrow c = 3 \times {10^8}\;m/s$
According to a theory formulated by Einstein, when a radiation of frequency $v$ is incident on a metal surface, it is absorbed in the form of discrete packets of energy called photons.This process involves the emission of light-generated electrons and a part of the energy $hv$ of a photon is used in removing an electron from the metal surface. The remaining energy is used in providing the emitted electrons with kinetic energy.Hence by Einstein’s photoelectric equation the equation obtained is as follows:
$K = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$
where, $K$ denotes the kinetic energy, $\lambda $ denotes the wavelength and ${\lambda _0}$ denotes the threshold wavelength.

Now, let us construct an equation relating the stopping potential and the kinetic energy of the ejected electrons. The stopping potential is the minimum value of negative potential applied to the anode of a photocell to make the photoelectric current equivalent to zero.The point where there are no electrons emitted is when the stopping potential is reached. The work done by the stopping potential on the fastest electron trying to escape the surface of the metal will be equivalent to its kinetic energy. Hence, the equation is as follows:
$K = e{V_0}$
The above equations are equated as they are similar (both are equivalent to kinetic energy).
We get:
$e{V_0} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$ --- (1)
where,
${V_0}$ is the stopping potential
When light of wavelength $n\lambda $ is incident on the metal, the stopping potential is found to be $\dfrac{{{V_0}}}{{n + 1}}$ as per the data given in the question. Using this we can construct another equation keeping equation ($1$) as a base. We get:
$\dfrac{{e{V_0}}}{{n + 1}} = \dfrac{{hc}}{{n\lambda }} - \dfrac{{hc}}{{{\lambda _0}}}$
By cross multiplying the terms and making this equation similar to equation ( $1$ ) we get:
$e{V_0} = (n + 1)\dfrac{{hc}}{{n\lambda }} - (n + 1)\dfrac{{hc}}{{{\lambda _0}}}$--- (2)
Now, we equate the two equations since their left hand part is the same. We get:
$\dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} = (n + 1)\dfrac{{hc}}{{n\lambda }} - (n + 1)\dfrac{{hc}}{{{\lambda _0}}}$

By taking out $hc$ as a common term and cancelling out this term on their side we get:
$\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}} = (n + 1)\dfrac{1}{{n\lambda }} - (n + 1)\dfrac{1}{{{\lambda _0}}}$
Expanding out the terms in brackets we get:
$\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}} = \dfrac{1}{\lambda } + \dfrac{1}{{n\lambda }} - \dfrac{n}{{{\lambda _0}}} - \dfrac{1}{{{\lambda _0}}}$
 The terms $\dfrac{1}{\lambda }$and $ - \dfrac{1}{{{\lambda _0}}}$ are taken out as common and cancelled. The resulting equation after rearranging the terms is:
$\dfrac{1}{{n\lambda }} = \dfrac{n}{{{\lambda _0}}}$
The aim here is to find the equation for threshold wavelength ${\lambda _0}$. Hence we by rearranging and cross multiplying the terms and making ${\lambda _0}$ the subject we obtain the final equation as:
${\lambda _0} = {n^2}\lambda $

Additional information: The phenomenon of emission of electrons from a metal surface when electromagnetic radiations of sufficiently high frequency are incident on it is known as photoelectric effect. Photons or light generated electrons are emitted out once the light is incident on the surface. The photoelectric effect depends on the intensity of light, the potential difference applied between the two electrons and the nature of the cathode material.

Note:The photoelectric effect takes place at only certain conditions. Different substances undergo photoelectric effect and emit photons only when exposed to radiations of different frequency and only when it surpasses the minimum required energy needed to emit photons.