Question

# The work function for metals A, B and C are respectively $1.92eV$, $2.0eV$ and $5eV$. According to Einstein’s equation, the metals which will emit photoelectrons for radiation of wavelength $4100A^{\circ}$ is/are:A. NoneB. ‘A’ onlyC. ‘A’ and ‘B’ onlyD. All three metals

Hint: Work function is inverse to the wavelength. If the work function increases, wavelength decreases, and if the work function decreases, the wavelength increases.

Formula used:
$E = h\nu$
Where, $E$=energy required to eject an electron,
$h$=planck’s constant which is equal to $6.626 \times {10^{ - 34}}{m^2}kg{s^{ - 1}}$,
$\nu$=frequency

Complete step by step solution:
Here we have to find the metals which can produce the radiation of wavelength $4100A^{\circ}$. There are $3$ metals $A, B$ and $C$. The work function of the metals $A, B$ and $C$ are $1.92eV, 2.0eV$, and $5eV$ respectively. Let’s find the wavelength of the three metals. $E = h\nu$ Here we have to find the wavelength. So, converting the formula into $E = \dfrac{{hc}}{\lambda }$. Now we can apply the known values.
$E = \dfrac{{hc}}{\lambda }$
We have a work function; we have to find the wavelength. So keeping the wavelength on one side, and other known values on another side.
$\lambda = \dfrac{{hc}}{E}$
Now the formula is ready to apply.
(i)To find the wavelength of metal $A$:
The work function for the metal $A$=$1.92eV$$\to 1.92 \times 1.6 \times {10^{ - 19}}V$. As $e$ is the charge of an electron.
$\lambda = \dfrac{{hc}}{E}$
On substituting the corresponding values,
$\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.92 \times 1.6 \times {{10}^{ - 19}}}}$
On simplifying the above equation, we get
$\Rightarrow \lambda = \dfrac{{6.626 \times 3}}{{1.92 \times 1.6}} \times {10^{ - 34 + 8 + 19}}$
$\Rightarrow \lambda = \dfrac{{19.878}}{{3.072}} \times {10^{ - 7}}$
On further simplification,
$\Rightarrow \lambda = 6.470 \times {10^{ - 7}}$
This shows that metal $A$ can emit photoelectrons of wavelength $4100A^{\circ}$. As the wavelength of metal $A$ is greater than the required wavelength.
(ii)To find the wavelength of metal $B$:
$E = \dfrac{{hc}}{\lambda }$
On substituting the corresponding values,
$\Rightarrow E = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2.0 \times 1.6 \times {{10}^{ - 19}}}}$
On simplifying the above equation, we get
$\Rightarrow \lambda = \dfrac{{19.878}}{{3.2}} \times {10^{ - 7}}$
$\Rightarrow \lambda = 6.211 \times {10^{ - 7}}$
On further simplification,
$\Rightarrow \lambda = 6211A^{\circ}$
From the wavelength of metal $B$, it can emit the photoelectrons of wavelength greater than $4100A^ {\circ}$. As it has a wavelength greater than the required wavelength.
(iii)To find the wavelength of metal $C$:
$\lambda = \dfrac{{hc}}{\lambda }$
On substituting the corresponding values,
$\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5 \times 1.6 \times {{10}^{ - 19}}}}$
On simplifying the above equation, we get
$\Rightarrow \lambda = \dfrac{{19.878}}{8} \times {10^{ - 7}}$
On further simplification,
$\Rightarrow \lambda = 2.484 \times {10^{ - 7}}$
$\Rightarrow \lambda = 2484A^{\circ}$
From the wavelength of metal $C$, it cannot emit photoelectrons of the wavelength of $4100A^{\circ}$. As it can produce wavelength up to $2484A^{\circ}$.

$\therefore$ Metals A and B can emit radiation of wavelength $4100A^{\circ}$. Hence, the correct answer is option (C).

Note:
When the electromagnetic radiation of frequency more than the threshold frequency, the photoelectric effect occurs. The minimum energy required to eject an electron from the metal surface is called the work function in the photoelectric effect.