and be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of

Question

$K_{1}$ and $K_{2}$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $λ_{1}$ and $λ_{2}$ , respectively. If $λ_{1} = 2 λ_{2}$ , then the work function of material is given by [2026]

Smart Achievers · Accepted Answer

(4)

$K_{1} = \frac{h c}{λ_{1}} - ϕ and K_{2} = \frac{h c}{λ_{2}} - ϕ$

$K_{2} = \frac{h c}{\frac{λ_{1}}{2}} - ϕ = 2 (\frac{h c}{λ_{1}}) - ϕ$

$K_{2} = 2 (K_{1} + ϕ) - ϕ = 2 K_{1} + ϕ$

$ϕ = K_{2} - 2 K_{1}$

Solution

Q.
$K_{1}$ and $K_{2}$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $λ_{1}$ and $λ_{2}$ , respectively. If $λ_{1} = 2 λ_{2}$ , then the work function of material is given by [2026]

1 $K_{2} + 2 K_{1}$

2 $2 K_{2} - K_{1}$

3 $K_{1} - 2 K_{2}$

4 $K_{2} - 2 K_{1}$

Ans.
(4)

$K_{1} = \frac{h c}{λ_{1}} - ϕ and K_{2} = \frac{h c}{λ_{2}} - ϕ$

$K_{2} = \frac{h c}{\frac{λ_{1}}{2}} - ϕ = 2 (\frac{h c}{λ_{1}}) - ϕ$

$K_{2} = 2 (K_{1} + ϕ) - ϕ = 2 K_{1} + ϕ$

$ϕ = K_{2} - 2 K_{1}$

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Solution

Q. K1 and K2 be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength λ1 and λ2, respectively. If λ1=2λ2, then the work function of material is given by [2026]

1 K2+2K1

2 2K2-K1

3 K1-2K2

4 K2-2K1

Ans. (4) K1=hcλ1-ϕ and K2=hcλ2-ϕ K2=hcλ12-ϕ=2(hcλ1)-ϕ K2=2(K1+ϕ)-ϕ=2K1+ϕ ϕ=K2-2K1