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प्रश्न
When a light of frequency 9 × 1014 Hz is incident on a metal surface, photoelectrons are emitted with a maximum speed of 8 × 105 ms−1. Determine the threshold frequency of the surface.
उत्तर
v = 9 × 1014 Hz
v = 8 × 105 ms−1
From Einstein’s photo electric equation
hv = hv0 – `1/2` mv2
hv0 = hv – `1/2` mv2
v0 = `("hv" - 1/2 "mv"^2)/"h"`
v0 = `"v" - ("mv"^2)/(2"h")`
= `9 × 10^14 - (9.1 xx 10^-31 xx 64 xx 10^10)/(2 xx 6.626 xx 10^-34)`
= `9 xx 10^14 - (582.4 xx 10^-21)/(13.252 xx 10^-34)`
= 9 × 1014 − 4.39 × 1014
= (9 − 4.39) × 1014
v0 = 4.61 × 1014 Hz
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