An isolated lead ball is charged upon continuous irradiation by EM radiation of wavelength `(lambda)=221` nm. The maximum potential attained by the lead ball, if its work function is `4.14 eV` , is
(Take `h = 6.63 xx 10^(-34) J. s, c = 3 xx 10^6 m//s, e=1.6 xx 10^(-19)`)

Options

A. 1.49 V
B. 2.67 V
C. 3.14 V
D. 0.51 V

Answer

A

Explanation

Given,
`lambda = 221 nm = 221 xx 10^(-9) m `
`phi = 4.14 e V`
`h = 6.63 xx 10^(-34) J.S`
`C = 3 xx 10^8` m/ sec
` e = 1.6 xx 10^(-18) C `
Maximum potential `(V_0) - (h/e) V- (phi_(0))/(e )`
`= h/e xx C /lambda - (phi_(0))/(e ) `
`= (6.63 xx 10^(-34) xx 3 xx 10^8 )/( 1.6 xx 10^(-19) xx 211 xx 10^(-9) ) - (4.14 xx 1.6 xx 10^(-19))/(1.6 xx 10^(-19))`
`= 1.485 V`
`V_(0) = 1.49 V`