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When a light of frequency 9 xx 10^(14)...

When a light of frequency `9 xx 10^(14) Hz` is incident on a metal surface, photoelectrons are emitted with a maximum speed of `8 xx 10^(5) ms^(-1)` . Determine the threshold frequency of the surface.

Text Solution

Verified by Experts

According to Einstein.s photoelectrc equation
`(1)/(2) mv_(max)^(2) = h(upsilon - upsilon_(0))`
`upsilon_(0) = upsilon - (mv_(max)^(2))/(2h)`
`= 9 xx 10^(14) - ((9.11 xx 10^(-31) xx (8 xx 10^(5))^(2))/(2. xx 6.6 xx 10^(-34)))`
`= 9 xx 10^(14) - ((583.04 xx 10^(-21))/(13.2 xx 10^(-34))) = 9 xx 10^(14) - 44.169 xx 10^(13)`
`= 9xx 10^(14) - 4.4 xx 10^(14)`
`upsilon_(0) = 4.6 xx 10^(14)` Hz
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