Electric field component of an EM radiation varies with time as E = a( ω₀ t + ω t ω₀ t), where ' a ' is a constant and ω = 10¹5} {sec}^{-1}, ω₀ = 5 × 10¹5} {sec}^{-1}. This radiation falls on a metal whose stopping potential is -2 {ev}. Then which of the following statement (s) is/are true ? (h = 6.62 × 10^{-34} {J}-{S})

For light of frequency ω, photoelectric effect is not possible

Stopping potential vs. frequency graph will be a straight line

The work function of the metal is –2 ev

The maximum kinetic energy of the photo electrons is 1.95 ev.

Test Index

WBJEE 2021 Physics and Chemistry Solved Paper

Category III (Q36 to Q40)

Carry 2 marks each and one or more option (s) is/are correct. If all correct answers are not marked and also no incorrect answer is marked then score = 2 × number of correct answers marked  actual number of correct answers. If any wrong option is marked or if any combination including a wrong option is marked, the answer will be considered wrong, but there is no negative marking for the same and zero mark will be awarded.

Question : 36 of 80

Marks: +1, -0

Electric field component of an EM radiation varies with time as

E = a(\cos \omega_{0} t + \sin \omega t \cos \omega_{0} t)

, where '

a

' is a constant and

\omega = 10^{15} \text{sec}^{-1}, \omega_{0} = 5 \times 10^{15} \text{sec}^{-1}

. This radiation falls on a metal whose stopping potential is

-2 \text{ev}

. Then which of the following statement (s) is/are true ?

(h = 6.62 \times 10^{-34} \text{J}-\text{S})

For light of frequency ω, photoelectric effect is not possible
Stopping potential vs. frequency graph will be a straight line
The work function of the metal is –2 ev
The maximum kinetic energy of the photo electrons is 1.95 ev.

Validate

Solution:

Stopping potential

= -2 \text{V} \Rightarrow \text{KE}_{\max} = 2 \text{ev}

(Option D incorrect)

E = a(\cos \omega_{0} t + \sin \omega t \cos \omega_{0} t)

= a \cos \omega_{0} t + \frac{a}{2} \{ \sin (\omega + \omega_{0}) t

+ \sin (\omega - \omega_{0}) t \}

Corresponding frequencies are

\frac{\omega_{0}}{\alpha \pi}, \frac{\omega + \omega_{0}}{\alpha \pi} \& \frac{\omega - \omega_{0}}{2 \pi}

\left(\frac{\omega + \omega_{0}}{2 \pi}\right)

determines maximum KE\ \&\ Stopping potential

Now Energy corresponding to

\frac{\omega + \omega_{0}}{2 \pi}

Corresponding energy

= h f

Stopping potential

= V_{0} = 2 \text{V} = 3.95 \text{ev}

Now

\text{KE}_{\max} = \text{ev}_{0} = h f - \phi \Rightarrow 2 \text{ev} = 3.95 \text{ev} - \phi

\Rightarrow \phi = 1.95 \text{ev}

for

w = 10^{15} \text{sec}^{-1}

frequency

= \frac{10^{15}}{2 \pi} = f_{1}

When frequency is

w, f_{1} = \frac{\omega}{2 \pi} = \frac{10^{15}}{2 \pi}

E = \frac{6.63 \times 10^{-34} \times 6 \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} \text{ev} = 3.95 \text{ev}

Energy

= h f_{1} = \frac{6.63 \times 10^{-34} \times 10^{15}}{2 \pi \times 1.6 \times 10^{-19}} \text{ev}

= 0.66 \text{v} < \phi

(work function)

Hence photoelectric effect not possible (option A correct)

Also

\text{ev}_{0} = h f - \phi

(Straight line) (option B correct)

\text{KE}_{\max} = E - \phi

\Rightarrow \phi = E - 2 \text{ev} = 1.95 \text{ev}

(option

c

is incorrect)

Go to Question:

Prev Question

Next Question