SELECT * FROM question_mgmt as q WHERE id=2780 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=125 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th physics Dual Nature Of Radiation And Matter light of frequency 7 21 times 1014 hz is inciden

Question:

Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Answer:

Frequency of the incident photon, v = 488 nm = 488 x 10-9 m

Maximum speed of the electrons, v = 6.0 × 105 m/s

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

Mass of an electron, m = 9.1 × 10−31 kg

For threshold frequency ν0, the relation for kinetic energy is written as: 1/2 mv2

= h(v - v0)

∴ v0 = v - mv2/2h

= 7.21 x 1014 - (9.1 x 10-31) x (6 x 105)2 / 2 x (6.626 x 10-34)

= 7.21 x 1014 - 2.472 x 1014

= 4.738 x 1014 Hz

Therefore, the threshold frequency for the photoemission of electrons is 4.738 × 1014 Hz.


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SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='8' AND ex_no!=0 AND status=1 and id!=2780 ORDER BY last_viewed_on desc limit 0,10

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