When photons of energy 4.0 eV fall on the surface of a metal A , the ejected photoelectrons have maximum kinetic energy TA (in eV ) and a de-Broglie wavelength λ A . When the same photons fall on the surface of another metal B , the maximum kinetic energy of ejected photoelectrons is TB=TA-1.5 eV . If the de-Broglie wavelength of these photoelectrons is λ B=2λ A , then the work function of metal B is

Question

When photons of energy 4.0 eV fall on the surface of a metal A , the ejected photoelectrons have maximum kinetic energy TA (in eV ) and a de-Broglie wavelength λ A . When the same photons fall on the surface of another metal B , the maximum kinetic energy of ejected photoelectrons is TB=TA-1.5 eV . If the de-Broglie wavelength of these photoelectrons is λ B=2λ A , then the work function of metal B is (A) 2eV (B) 3eV (C) 2.5eV (D) 3.5eV

Tardigrade Admin · Accepted Answer

The relation between de-Broglie wavelength and kinetic energy is

λ = \frac{h}{2 K m _{e}}

\frac{λ _{A}}{λ _{B}} = \frac{K _{B}}{K _{A}} \Rightarrow \frac{1}{2} = \frac{T _{A} - 1.5}{T _{A}} \Rightarrow T_{A} = 2 e V

\Rightarrow K_{B} = 2 - 1.5 = 0.5 e V

So, the work-function of

B

is

ϕ_{B} = 4.0 - 0.5 = 3.5 e V

$2 e V$

$3 e V$

$2.5 e V$

$3.5 e V$

2eV

3eV

2.5eV

3.5eV

The relation between de-Broglie wavelength and kinetic energy is λ=2Kme​​h​ λB​λA​​=KA​​KB​​​⇒21​=TA​TA​−1.5​​⇒TA​=2eV ⇒KB​=2−1.5=0.5eV So, the work-function of B is ϕB​=4.0−0.5=3.5eV

$2 e V$

$3 e V$

$2.5 e V$

$3.5 e V$