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When a certain metal was irradiated wit...

When a certain metal was irradiated with light of frequency `3.2 xx 10^(16)s^(-1)` the photoelectrons emitted had twice the KE as did photoelectrons emitted when the same metal was irradiated with light of frequency`2.0 xx 10^(16)s^(-1)` . Calculate the thereshold frequency of the metal.

Text Solution

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Kinetic energy of emitted photelectrons is
K.E. `= hv - hv - hv_@ = h(v - v_@)`
For the light of frequency `3.2 xx 10^(16) "sec"^(-1)`
`K.E_1 = h(3.2 xx 10^(16) - v_@)`
For the light of frequency `2 xx 10^(16) "sec"^(-1)`
`KE_2 = h(2.0 xx 10^(16) - v_@)`
It is given that `KE_1 = 2KE_2`
`:. h (3.2 xx 10^(16) - v_@) = 2h (2.0 xx 10^(16) - v_@)`
`3.2 xx 10^(16) - v_@ = 2 (2.0 xx 10^(16) - v_@)`
`3.2 xx 10^(16) - v_@ = 4.0 xx 10^(16) - 2v_@`
`-v_@ + 2v_@ = (4.0 - 3.2) xx 10^(16)`
or `" " v_@ = 0.8 xx 10^(16)`
or `" " = 8.0 xx 10^(15) "sec"^(-1)`
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