Surface of a certain metal is first illuminated with light of wavelength ${\lambda _1} = 350nm$ and then, by light of wavelength ${\lambda _2} = 540nm$. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to:
(Energy of photon $ = \dfrac{{1240}}{{\lambda ({\text{in nm)}}}}eV$)
(A) $1.8$
(B) $1.4$
(C) $2.5$
(D) $5.6$

Hint
From Einstein's equation of photoelectric effect we can write the equations for the light of 2 different wavelengths in the 2 given cases. By comparing the kinetic energies in the 2 cases we can find the work function of the metal.
In this solution we will be using the following formula,
$\Rightarrow E = \phi + K{E_e}$
where $E$ is the energy of the photons,
$\phi $ is the work function of the metal and
$K{E_e}$ is the kinetic energy of the electrons.

Complete step by step answer
From Einstein's equation of photoelectric effect, the energy of the photons provides the work function of the metal and the kinetic energy of the electrons. It is written as,
$\Rightarrow E = \phi + K{E_e}$
Now the energy of the photons are given by the formula,
$\Rightarrow E = \dfrac{{hc}}{\lambda }$ where $\lambda $ is the wavelength.
And the kinetic energy of the electrons can be written as,
$\Rightarrow \dfrac{{hc}}{{{\lambda _1}}} = \phi + \dfrac{1}{2}{m_e}v_e^2$
Since the work function of the metal is the characteristic property of the metal, it will be the same in both cases.
We can write the equation in the 2 cases provided in the question as,
$\Rightarrow \dfrac{{hc}}{{{\lambda _1}}} = \phi + \dfrac{1}{2}{m_e}{\left( {2{v_e}} \right)^2}$ and
$\Rightarrow \dfrac{{hc}}{{{\lambda _2}}} = \phi + \dfrac{1}{2}{m_e}v_e^2$
Since it is given in the question, the velocity in the first case is twice the second.
Therefore we can write the equations as,
$\Rightarrow 4 \times \dfrac{1}{2}{m_e}v_e^2 = \dfrac{{hc}}{{{\lambda _1}}} - \phi $
and $\dfrac{1}{2}{m_e}v_e^2 = \dfrac{{hc}}{{{\lambda _2}}} - \phi $
So we can equate the second equation in the first and get,
$\Rightarrow 4 \times \left( {\dfrac{{hc}}{{{\lambda _2}}} - \phi } \right) = \dfrac{{hc}}{{{\lambda _1}}} - \phi $
On opening the brackets and taking the like terms to one side we get,
$\Rightarrow \dfrac{{4hc}}{{{\lambda _2}}} - \dfrac{{hc}}{{{\lambda _1}}} = - \phi + 4\phi $
Simplifying we get,
$\Rightarrow hc\left( {\dfrac{4}{{{\lambda _2}}} - \dfrac{1}{{{\lambda _1}}}} \right) = 3\phi $
So the work function is given by,
$\Rightarrow \phi = \dfrac{{hc}}{3}\left( {\dfrac{4}{{{\lambda _2}}} - \dfrac{1}{{{\lambda _1}}}} \right)$
In the question we are provided, ${\lambda _1} = 350nm$ and ${\lambda _2} = 540nm$ and,
Energy of photon $ = \dfrac{{1240}}{{\lambda ({\text{in nm)}}}}eV$
So substituting we get
$\Rightarrow \phi = \dfrac{{1240}}{3} \times \left( {\dfrac{4}{{540}} - \dfrac{1}{{350}}} \right)$
On doing the calculation of both the terms we get
$\Rightarrow \phi = \dfrac{{1240}}{3}\left( {\dfrac{{1400 - 540}}{{350 \times 540}}} \right)$
On doing the calculation inside the bracket we get
$\Rightarrow \phi = \dfrac{{1240}}{3} \times 4.5 \times {10^{ - 3}}$
The calculation gives the work function as,
$\Rightarrow \phi = 1.8eV$
So the correct answer is option (A) $1.8eV$.

Note
The phenomenon of the emission of electrons from a metal surface when an electromagnetic radiation falls on the surface of the metal is called the photoelectric effect. The kinetic energy of the emitted photons depends on the intensity of the electromagnetic radiation.