Question

Work function of a metal is 2.51 eV. Its threshold frequency is:
A. \[5.9 \times {10^{14}}\,cycle/\sec \]
B. \[6.5 \times {10^{14}}\,cycle/\sec \]
C. \[9.4 \times {10^{14}}\,cycle/\sec \]
D. \[6.08 \times {10^{14}}\,cycle/\sec \]

Answer 1

Hint: The equation for photoelectric effect is\[h\nu = {\phi _o} + {E_k}\].The minimum amount of energy required for electron emission from the metal surface is the work function of that metal and the frequency of light corresponding to this minimum energy is called threshold frequency and the corresponding wavelength is called threshold wavelength. It is calculated as work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _0}}}\]. In electron volts, this work function is equal to stopping potential. The work function needs to be converted to joules from eV to determine the threshold wavelength or the frequency.

Formula(e) Used:
Energy, \[E = h\nu = \dfrac{{hc}}{\lambda }\]
Equation for photoelectric effect, \[h\nu = {\phi _o} + {E_k}\]
Work function, \[{\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}}\]
Where,
E = energy of incident radiation
\[{\phi _o}\]= Work function of the metal
\[{E_k}\]= energy of the emitted photoelectron
\[c{\rm{ }} = \text{speed of light} = 3 \times {10^8}m/s\]
\[\nu \]= frequency of the light,
\[{\nu _o}\]= threshold frequency,
\[\lambda \]= wavelength of the light,
\[{\lambda _o}\]= threshold wavelength
\[e = 1.6 \times {10^{ - 19}}C\]= electronic charge

Complete step by step solution:
Work function thus can be defined as the minimum energy required for photoelectron emission from the surface of a particular metal.

Given: In this photoelectric experiment, the work function of the metal is given to be \[2.51eV\]. We need to determine the corresponding threshold frequency for this metal.
Equation for photoelectric effect is,
\[h\nu = {\phi _o} + {E_k} \\ \]---- (1)
\[\Rightarrow {\phi _o} = h{\nu _o} = \dfrac{{hc}}{{{\lambda _o}}} \\ \]---(2)
We know that \[1eV = 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow {\phi _o} = 2.51\,eV\]
\[\Rightarrow {\phi _o} = 2.51 \times 1.6 \times {10^{ - 19}}J\]

From equation (2), threshold frequency can be calculated as
\[{\nu _o} = \dfrac{{{\phi _o}}}{h} \\ \]---- (3)
Substituting all numerical values in equation (3),
\[{\nu _o} = \dfrac{{2.51 \times 1.6 \times {{10}^{ - 19}}}}{{6.64 \times {{10}^{ - 34}}}}\,cycle/\sec \\ \]
\[\therefore {\nu _o} = 6.08 \times {10^{14}}\,cycle/\sec \]
Therefore, the threshold frequency is \[{\nu _o} = 6.08 \times {10^{14}}\,cycle/\sec \].

Hence option D is the correct answer.

Note: No emission of electrons or any photoelectric effect from the metal will take place if the energy supplied is less than the work function. To determine the corresponding threshold wavelength, we use the formula, \[{\lambda _o} = \dfrac{c}{{{\nu _o}}}\]. Work function is a characteristic property of the metal and is independent of incident or emission wavelength or frequency or intensity of radiation and any such external factors. Work function must be in joules to determine the threshold wavelength or threshold frequency. In order to calculate stoppage potential , the unit of work function should be electron-volts.