Published by:
CGP EDU Academic Team
Published on: August 14, 2026
A light of wavelength 3000 
falls on a metal surface. Ejected e~ is further accelerated by a potential difference of 2V , then final K.E of the e~ is found to be 8 x 10 -19 J . If threshold energy for the metal surface is 
eV . Then find the numerical value of 
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
9.00
KE final = E photon + E potential 
where ( 
) = threshold energy for metal surface
Hence, 
= (4.12 + 2 - 5) eV = 1.125 eV
Numerical value = 8 x 1.125 = 9
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