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When light of wavelength 248 nm falls on...

When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is `"_________"Å` (Round off to the Nearest Integer).
[Use `: sqrt3 = 1.73 , h - 6.63 xx 10 ^(-34) Js`
`m _(c ) = 9.1 xx 10 ^(-31) kg , c =3.0 xx 10 ^(8) ms ^(-1) ,1eV = 1. 6 xx 10 ^(-19))]`

Text Solution

Verified by Experts

The correct Answer is:
9
Energy incident `=(hc)/lamda`
`=(6.63 xx 10^(-34) xx 3.0xx10^8)/(248 xx 10^(-9) xx1.6xx10^(-19))eV`
`=(6.63xx3xx100)/(248 xx1.6)`
` = 0.05 eV xx100 = 5eV`
Now using
`E =phi +` K.E.
5 = 3 + K.E.
K.E. = 2eV =`3.2 xx10^(-19) J`
For de-Broglie wavelength `lamda = h/(mv)`
K.E. `=1/2mv^2`
so `v=sqrt((2KE)/m)`
hence `lamda=h/(sqrt(2KExxm))`
`=(6.63xx10^(-34))/(sqrt(2xx3.2xx10^(-19) xx9.1xx10^(-31))`
`=( 6.63)/(7.6) xx(10^(-34))/(10^(-25))=(66.3xx10^(-10) m)/(7.6)`
`= 8.72xx10^(-10)m`
`=9xx10^(-10)m = 9Å`
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