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A monochromatic neon lamp with wavelengt...

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm ?

A

0.96V

B

1.5V

C

1.25V

D

0.24 V

Text Solution

Verified by Experts

The correct Answer is:
C
`kE_("max") = (hc)/(lambda) + phi`
Or ` eV_(0) = (hc)/( lambda_(1)) + phi `
When ` lambda _(i) = 670 . 5 nm , V_(0)= 0 . 48`
When `lambda_(i) = 474 .6 nm , V_(0) = ?`
So, ` e (0.48) = (1240)/(670.5) + phi ` . . . (i)
`e (V_(0)) = (1240)/( 474.6) + phi` . . . (ii)
(ii) - (i)
`e (V_(0) - 0.48) = 1240 ((1)/( 474.6) - (1)/(670.5)) eV`
`V_(0) = 0.48 + 1240 (( 670 . 5 - 474 . 6)/(474 . 6 xx 670.5)) ` Volts
`V_(0) = 0.48 + 0.76`
`V_(0) = 1.24 V~~ 1.25 V `
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