Photons having energy 1 eV and 2.5 eV successively incident on a metal, having work function is 0.5 eV. The ratio of maximum speed of emitted electrons is
A. 1:2
B. 2:1
C. 3:1
D. 1:3

Hint: First take the Einstein photoelectric equation, Then write that formula in terms of energy, and further as in terms of wavelength, planck constant and speed of light. Now put the value of both the energies of protons in two different equations and then find the solution. Then equate both the results to find the ratio between velocity of both the electrons.

Formula used:
$hv=\phi +\dfrac{1}{2}m{{v}^{2}}$
$E-\phi =\dfrac{1}{2}m{{v}^{2}}$
$\dfrac{hc}{\lambda }-\phi =\dfrac{1}{2}m{{v}^{2}}$
$E=\dfrac{hc}{\lambda }$

Complete step-by-step answer:
We know that according to Einstein Photoelectric equation we have,
$hv=\phi +\dfrac{1}{2}m{{v}^{2}}$ ,
The equation given by Einstein is,
$E-\phi =\dfrac{1}{2}m{{v}^{2}}$.
We can also represent this as,
$\dfrac{hc}{\lambda }-\phi =\dfrac{1}{2}m{{v}^{2}}$.( as $E=\dfrac{hc}{\lambda }$)
Now in case of photon having energy of 1eV,
$\left( 1-0.5 \right)eV=\dfrac{1}{2}m{{v}^{2}}$ ,
$0.5eV=\dfrac{1}{2}m{{v}^{2}}$…… this is equation 1.
Now similarly if the second phone has the energy of 2.5 eV,
$\left( 2.5-0.5 \right)eV=\dfrac{1}{2}m{{{v}'}^{2}}$,
$2eV=\dfrac{1}{2}m{{{v}'}^{2}}$……….. this is equation 2.
Now on dividing equation 2 by equation 1,
$\dfrac{2eV}{0.5eV}=\dfrac{\dfrac{1}{2}m{{{{v}'}}^{2}}}{\dfrac{1}{2}m{{v}^{2}}}$ ,
\[\dfrac{2eV}{0.5eV}=\dfrac{{{{{v}'}}^{2}}}{{{v}^{2}}}\],
\[\dfrac{{{v}'}}{v}=\sqrt{4}\],
\[\dfrac{{{v}'}}{v}=2\],
\[\dfrac{{{v}'}}{v}=\dfrac{2}{1}\],

So, the correct answer is “Option B”.

Additional Information: The smallest discrete amount is known as a photon, it is the basic unit of all light.
Photons have a zero rest mass.
Work function of a metal is the minimum work done or the energy required thermodynamically by a metal substance to emit a single electron from its body.
The energy carried by a single photon is known as photon energy.

Note: In the equation $hv=\phi +\dfrac{1}{2}m{{v}^{2}}$, $\phi $ is the work function, $\dfrac{1}{2}m{{v}^{2}}$ is the kinetic energy of the particle having mass and velocity represented by ‘m’ and ‘v’ , respectively. ‘h’ is the planck's constant. And in this$E=\dfrac{hc}{\lambda }$ formula, $\lambda $ is the wavelength of the photon rays.