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Solution - The Threshold Frequency for a Certain Metal is 3.3 × 1014 Hz. If Light of Frequency 8.2 × 1014 Hz is Incident on the Metal, Predict the Cutoff Voltage for the Photoelectric Emission. - CBSE (Commerce) Class 12 - Physics

ConceptPhotoelectric Effect and Wave Theory of Light

Question

The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Solution

Threshold frequency of the metal, `v_0 = 3.3 xx 10^(14) Hz`

Frequency of light incident on the metal, `v = 8.2 xx 10^14 ` Hz

Charge on an electron, e = 1.6 × 10−19 C

Planck’s constant, h = 6.626 × 10−34 Js

Cut-off voltage for the photoelectric emission from the metal = `V_0`

The equation for the cut-off energy is given as:

`eV_0 = h(v -v_0)`

`V_0 = (h(v - v_0))/e`

`= (6.626 xx 10^(-34) xx (8.2 xx 10^14 - 3.3 xx 10^(14)))/(1.6 xx 10^(-19)) = 2.0292 V` 

Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V

Is there an error in this question or solution?

APPEARS IN

 NCERT Physics Textbook for Class 12 Part 2 (with solutions)
Chapter 11: Dual Nature of Radiation and Matter
Q: 8 | Page no. 408

Reference Material

Solution for question: The Threshold Frequency for a Certain Metal is 3.3 × 1014 Hz. If Light of Frequency 8.2 × 1014 Hz is Incident on the Metal, Predict the Cutoff Voltage for the Photoelectric Emission. concept: null - Photoelectric Effect and Wave Theory of Light. For the courses CBSE (Commerce), CBSE (Arts), CBSE (Science)
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