
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
161.1k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE
