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The photoelectric threshold wavelength o...

The photoelectric threshold wavelength of silver is `3250 xx 10^(-10) m`. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength `2536 xx 10^(-10) m` is
`(Given h = 4.14 xx 10^(6) ms^(-1) eVs` and `c = 3 xx 10^(8) ms^(-1))`

A

`~~6xx10^(5) ms^(-1)`

B

`~~0.6xx10^(6) ms^(-1)`

C

`~~61xx10^(3) ms^(-1)`

D

`~~0.3xx10^(6) ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
`lambda_(0)=3250xx10^(-10)m`
` lambda=2536xx10^(-10)m`
`phi=(1242 eV-nm)/(325nm)=3.82 eV`
`hv=(1242 eV-nm)/(253.6 nm)=4.89 eV`
`KE_("max")=(4.89-3.82)eV=1.077eV`
`(1)/(2) mv^(2)=1.077xx1.6xx10^(-19)`
`v=sqrt((2xx1.077xx1.6xx10^(-19))/(9.1xx10^(-31)))`
`v=0.6xx10^(6)m//s `
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