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When a metal plate is exposed to light of wavelength 400 nm, a negative potential of 1.1V is found to be needed to stop the photo current. The threshold wavelength of the photoelectrons is--------nm.

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answer is 620.

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Detailed Solution

Stopping potential

$1.1 Vi.e.,$

$K_{max} = 1.1 eV h v = w_{0} + K E_{max} λ = 4000 A^{0}$

$\frac{12400}{4000} = w_{0} + 1.1 eV$

$w_{0} = 2 eV λ_{0} = \frac{12400}{2} = 6200 A^{0} = 620 nm$

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