Question

A 20% increase in the energy of the incident radiation causes the kinetic energy of the photoelectron emitted from the metal surface to increase from 0.5 eV to 0.8 eV. Find the work function of the metal.
A) 0.65 eV
B) 1 eV
C) 1.3 eV
D) 1.5 eV

Answer 1

Hint: Work function ${\varphi }_0$ refers to the minimum energy needed for the electron to escape from the metal surface. The electrons thus emitted will have maximum kinetic energy.

Formula used:
The maximum kinetic energy of an emitted photoelectron is given by, $K_{max}=E-{\varphi }_0$ where $E$ is the energy of the incident radiation and ${\varphi }_0$ is the work function of the metal.

Complete step by step solution:
Step 1: List the data provided in the question.
When radiation of energy $E$ is incident on the metal surface, the kinetic energy of the emitted photoelectron is $K_{max}=0.5\text{eV}$ .
When the energy of the incident radiation increases the kinetic energy of the emitted electron is $K{}_{max}^{\prime }=0.8\text{eV}$ .
Step 2: Express the new energy (increased energy) $E^{\prime }$ in terms of the initial energy $E$.
It is given that the energy of the incident radiation is increased by 20%.
i.e., the new energy is $E^{\prime }=E+0.2E=1.2E$ .
Step 3: Express the photoelectric equation for the two cases (initial and new).
The photoelectric equation for the initial case is given by, $K_{max}=E-{\varphi }_0$ .
The photoelectric equation for the increase in incident energy is given by,$K{}_{max}^{\prime }=E^{\prime }-{\varphi }_0=1.2E-{\varphi }_0$ .
The work function ${\varphi }_0$ is property of the metal surface and therefore does not vary with change in incident energy.
Substitute values for $K_{max}=0.5\text{eV}$ , $K{}_{max}^{\prime }=0.8\text{eV}$ in the corresponding equations.
Then the above equations become, $0.5\text{eV}=E-{\varphi }_0$ ---- (1) and $0.8\text{eV}1.2E-{\varphi }_0$ ----- (2).
Step 4: Solve equations (1) and (2) to find the work function.
Equation (1) is $0.5\text{eV}=E-{\varphi }_0$ .
Equation (2) is $0.8\text{eV = }1.2E-{\varphi }_0$ .
Multiplying equation (1) with 1.2 gives $\text{0}\text{.6eV}=1.2E-1.2{\varphi }_0$ ----- (3).
Subtracting equation (3) from (2), we get $\left(\text{0}\text{.8}-\text{0}\text{.6}\right)\text{eV}=\left(1.2-1.2\right)E-\left(1-1.2\right){\varphi }_0$ .
i.e., $\text{0}\text{.2eV}=0.2{\varphi }_0$ .
On solving the above equation we get the work function of the metal surface ${\varphi }_0=1\text{eV}$ .

$\therefore$ The work function of the metal surface $=1\text{eV}$. Hence, Option (B) is correct.

Note:
The energy of the incident radiation is $E=h\nu$ where $h$ is the Plank’s constant and $\nu$ is the frequency of the incident radiation. A change in energy effectively means that the frequency of the incident radiation is changed. An increase in incident energy leads to an increase in the kinetic energy of the emitted electron.