For intensity I of a light of wavelength 5000 overset° A the photoelectron saturation is 0.40μ A and stopping potential is 1.36V , the work function of metal is

Question

For intensity I of a light of wavelength 5000 overset° A the photoelectron saturation is 0.40μ A and stopping potential is 1.36V , the work function of metal is (A) 2.47eV (B) 0.36eV (C) 1.10eV (D) 0.43eV

Tardigrade Admin · Accepted Answer

By using E=W0+Kmax E=(12375/5000)=2.475eV and Kmax=eV0=1.36eV So 2.475=W0+1.36⇒ W0=1.1eV

Q. For intensity $I$ of a light of wavelength $5000 A^{\circ}$ the photoelectron saturation is $0.40 μ A$ and stopping potential is $1.36 V$ , the work function of metal is

$2.47 e V$

$0.36 e V$

$1.10 e V$

$0.43 e V$

By using $E = W_{0} + K_{ma x}$
$E = \frac{12375}{5000} = 2.475 e V$ and $K_{ma x} = e V_{0} = 1.36 e V$
So $2.475 = W_{0} + 1.36 \Rightarrow W_{0} = 1.1 e V$

Q. For intensity I of a light of wavelength 5000A∘ the photoelectron saturation is 0.40μA and stopping potential is 1.36V , the work function of metal is

2.47eV

0.36eV

1.10eV

0.43eV