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 funtion is 4. 14 eV is (take, h `=6.63xx10^(-34)` J-s, `c=3xx 10^(8)m//s,e=1.6xx10^(-19)C`)

Options

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

Answer

A

Explanation

Given, `lambda=2.21 xx 10 ^(-7)m`
So, energy of imparted by radiation in the lead ball,
`E = (hc)/(lambda)= (6.62xx 10^(34)xx3xx10^(8))/(2.21xx10^(-7))= 9 xx 10^(-19)J`
or `E=(9xx10^(-19))/(1.6xx10^(19))eV = 563 ` eV
Work funtion of lead ball == 4. 14 eV
So, maximum energy attained by ball is
5.63-4.14 = 1.49 eV
Equlvalent maximum potential is 1.49 V.