Find the difference of kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500Å and 5000Å. `h=6.62xx10^(-34)Js`.

Question

Find the difference of kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500Å and 5000Å. `h=6.62xx10^(-34)Js`.

Find the difference of kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500Å and 5000Å. `h=6.62xx10^(-34)Js`.
A. 1.61 eV
B. 2.47 eV
C. 3.96 eV
D. `3.96xx10^(-19)eV`

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
As, `DeltaE=(hc)/(lamda_(1))-(hc)/(lamda_(2))=(hc(lamda_(2)-lamda_(1)))/(lamda_(1)lamda_(2))` (in eV)
`=(6.62xx10^(-34)xx3xx10^(8)xx(5000-2500)xx10^(-10))/(2500xx5000xx10^(-20)xx1.6xx10^(-19))`
`=2.47eV`

5 views Answered 2023-02-05 15:29