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A metal is irradiated with light of wave...

A metal is irradiated with light of wavelength 600 nm. Given that the work function of the metal is 1.0 eV, the de Broglie wavelength of the ejected electron is close to:

A

`6.6xx10^(-7)m`

B

`8.9xx10^(-11)m`

C

`1.19xx10^(-9)m`

D

`6.6xx10^(-13)m`

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("absorbed") =(hc)/(lambda)=(6.626xx10^(-34)Js xx 3xx10^(8)ms^(-1))/(600xx10^(-9)m)=0.0331xx10^(-17)J`
`E_("absorbed")=E_(0)+KE, KE=E_("absorbed")-E_(0)=3.31xx10^(-19)J-1.6 xx 10^(-19) J = 1.71 xx 10^(-19)J`
de Broglie wavelength `lambda=(h)/(sqrt(2Em))=(6.626xx10^(-34)Js)/(sqrt(2xx9.1xx10^(-31)kg xx 1.71 xx 10^(-19) J))=1.188xx10^(-9)m`
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