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Light of wavelength `lambda` falls on metal having work functions `hc//lambda_(0)` . Photoelectric effect will take place only if :

A

`lamda ge lamda_(0)`

B

`lamda ge 2lamda_(0)`

C

`lamda le lamda_(0)`

D

`lamda = 4lamda_(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
According to Einstein.s photoelectric equation
`K = (hc)/lamda - (hc)/(lamda_0)`
K is positive , i.e, photoelectric emission will take place if
`(hc)/lamda ge (hc)/lamda_0 " or " lamda le lamda_0`
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