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A metal is irradiated with light of wave...

A metal is irradiated with light of wavelength 660 nm. Given that the work function of the metal is `1.0eV,` the de Broglie wavelength of the ejected electron is close to-

A

`6.6xx10^(-7)m`

B

`8.9xx10^(-11)m`

C

`1.3xx10^(-9)m`

D

`6.6xx10^(-13)m`

Text Solution

Verified by Experts

The correct Answer is:
C
`E=phi+KE.`
`because E=(hC)/(lambda)=(6.6xx10^(-34)xx3xx10^(8))/(660xx10^(-9))`
`=3xx10^(-19)J`
`phi="lev"=1.6xx10^(-19)J`
K.E. `=3xx10^(-19)-1.6xx10^(-19)=1.4xx10^(-19)J`
for wave length of emitted electron
`lambda=(h)/(sqrt(2mKE))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.4xx10^(-19)))=(6.6xx10^(-9))/(5xx10^(-25))=1.32xx10^(-9)"meter"`
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