When light of wavelength `lambda` is incident on photosensitive surface, the stopping potential is V. When light of wavelength `3lambda` is incident o

Question

When light of wavelength `lambda` is incident on photosensitive surface, the stopping potential is V. When light of wavelength `3lambda` is incident o

When light of wavelength `lambda` is incident on photosensitive surface, the stopping potential is V. When light of wavelength `3lambda` is incident on same surface, the stopping potential is `V/6`Thereshould wave length for the surface is
A. `2lamda`
B. `3lamda`
C. `4 lamda`
D. `5lamda`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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

Correct Answer - D
`W_(0)=(hc)/(lamda)-eV_(s)`
`W_(0)=(hc)/(lamda_(1))-eV_(S_(1))=(hc)/(lamda_(2))-eV_(S_(2))`
`:.(hc)/(lamda)-eV=(hc)/(3lamda)-(eV)/(6)`
`(hc)/(lamda)-(hc)/(3lamda)=eV-(eV)/(6)`
`(hc)/(lamda)(1-(1)/(3))=(5eV)/(6)`
`(hc)/(lamda)((3-1)/(3))=(5eV)/(6)`
`(hc)/(lamda)=(5eV)/(6)xx(3)/(2)=(5eV)/(4):.eV=(4hc)/(5lamda)`
`:.W_(0)=(hc)/(lamda)-eV_(S_(1))`
`W_(0)=(hc)/(lamda)-(4hc)/(5lamda)`
`(hc)/(lamda_(0))=hc((1)/(lamda)-(4)/(5lamda))`
`(1)/(lamda_(0))=(5lamda-4lamda)/(5lamda^(2))`
`(1)/(lamda_(0))=(lamda)/(5lamda^(2))`
`(1)/(lamda_(0))=(1)/(5lamda)`
`:.lamda_(0)=5lamda`