Q.

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 3315 Å is 0.21 V. If the threshold frequency is   x  × 1013 / s, where  x is ____

(Given, speed light = 3 × 108 m/s, Planck’s constant = 6.63 × 10–34 Js)

[Round off to two decimal places]

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

answer is 85.43.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

hν=hν0+K.Emax124003315=hν0+0.21hν0=3.740.21=3.53 eVν0=3.53×1.6×10196.63×1034=85.43×1013HzX=85.43

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon