Question

In an experiment on photoelectric effect photons of wavelength 300 nm eject electrons from a metal of work function 2.25 eV. A photon of energy equal to that of the most energetic electron corresponds to the following transition in the hydrogen atom.
A. n=2 to n=1 state
B. n=3 to n=1 state
C. n=3 to n=2 state
D. n=4 to n=3 state

Answer 1

Hint: When light strikes a metal surface, the photoelectric effect occurs and electrons are expelled from the metal's surface. The value of the work function, which is dependent on the metal, indicates the minimum amount of energy needed to cause photoemission of electrons from a metal surface.

Formula used:
Here, we may express it as follows:
$\dfrac{{hc}}{\lambda } = \phi + K.E$.
where K.E. refers to the electron's kinetic energy at emission and E represents the photon's energy upon impact.
The energy of the nth Bohr orbit is given by,
$E = - \dfrac{{13.6}}{{{n^2}}}eV$

Complete step by step solution:
First let’s know about the photoelectric effect. In the photoelectric effect, light is incident on a metal surface, which causes electrons to be expelled from its surface. Photoelectrons are the name given to these expelled electrons. Note that the frequency of the light that is incident on the metal's surface affects both the emission of photoelectrons and the kinetic energy of the photoelectrons that are expelled.

The term "photoemission" refers to the method by which photoelectrons are released from the metal surface as a result of light. The photoelectric effect happens because electrons at the metal's surface have a tendency to absorb energy from the light that strikes them and use it to defy the forces that pull them toward the metallic nucleus. Below is a diagram that explains how the photoelectric effect causes photoelectrons to be emitted.

The value of the work function, $\phi $, which is dependent on the metal, is the minimum amount of energy needed to cause photoemission of electrons from a metal surface. Rest of the energy converted to kinetic energy of the electron. We can formulate it as $E = \phi + K.E.$ here E is the energy of the incident photon and K.E. is the kinetic energy of an emitted electron.

It is also written as $\dfrac{{hc}}{\lambda } = \phi + K.E$ where h=Planck constant and c is the speed of light and $\lambda $ is the wavelength of light incident. Now if we put the given values then we get
$\dfrac{{hc}}{\lambda } = \phi + K.E$
$\Rightarrow K.E. = \dfrac{{hc}}{\lambda } - \phi = \dfrac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{300 \times {{10}^{ - 9}} \times 1.6 \times {{10}^{ - 19}}}}eV - 2.25\,eV \\ $
$\Rightarrow K.E. = (4.14 - 2.25)\,eV = 1.89\,eV $

From Bohr model we can evaluate write,
$E = - \dfrac{{13.6}}{{{n^2}}}eV$
Here n is the orbit number. So, in the option two orbits are given and we have to find out the energy difference between the orbits.
When n=1,
${E_1} = - 13.6\,eV$
When n=2,
${E_2} = - 3.4\,eV$
When n=3,
${E_3} = - 1.5\,eV$
When n=4,
${E_4} = - 0.85\,e{V_{}}$
From here we observe that if we do ${E_2} - {E_3} = 1.9\,eV$. Which matches with the K.E. of the emitted electron. So, the transition of the hydrogen atom happens from n=3 to n=2.

Therefore, option C is the correct answer.

Notes In the photoelectric effect, when light strikes a metal surface, electrons are ejected from the metal's surface. The lowest amount of energy required to trigger photoemission of electrons from a metal surface is given by the value of the work function, which depends on the metal. The remaining energy was transformed to the electron's kinetic energy. Here, we may formulate it by $E = \phi + K.E.$ . where, E stands for the photon's energy at impact, and K.E. for the electron's kinetic energy upon emission. Energy of the nth orbit of Bohr’s orbit is given by $E = - \dfrac{{13.6}}{{{n^2}}}eV$ .