Light of wavelength `lambda_(A)` and `lambda_(B)` falls on two identical metal plates A and B respectively . The maximum kinetic energy of photoelectrons in `K_(A)` and `K_(B)` respectively , then which one of the following relations is true ? `(lambda_(A) = 2lambda_(B))`

Options

A. `K_(A) lt (K_(B))/(2)`
B. `2K_(A)=K_(B)`
C. `K_(A)=2K_(B)`
D. `K_(A) gt 2K_(B)`

Answer

A

Explanation

From photoelectric equation `hv=(hC)/(lambda)=W_(0)+K_("max")`
`:.(hC)/(lambda_(A))=W_(0)+K_(A)`
`:.(hC)/(2lambda_(B))=W_(0)+K_(A) ( :'lambda_(A)=2lambda_(B))` ...(1)
and `(hC)/(lambda_(B))=W_(0)+K_(B)` ...(2)
Subtracting (1) from (2), we get
`hC[(1)/(lambda_(B))-(1)/(2lambda_(B))]=K_(B)-K_(A)`
`:.(hC)/(2lambda_(B))=K_(B)-K_(A)`
`:.(hC)/(2lambda_(B))=2K_(B)-2K_(A)` ... (3)
From (2) and (3),
`2K_(B)-2K_(A)=W_(0)+K_(B)`
`K_(B)-2K_(A)=W_(0) " " :.2K_(A)=K_(B)-W_(0)`
`:.K_(A)=(K_(B)-W_(0))/(2) :.K_(A) lt (K_(B))/(2)`