A photo-emissive substance is illuminated with a radiation of wavelength so that it releases electrons with de-Broglie wavelength . The longest wavelength of radiation that can emit photoelectron is . Expression for de-Broglie wavelength is given by: (m

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Q.
A photo-emissive substance is illuminated with a radiation of wavelength $λ_{i}$ so that it releases electrons with de-Broglie wavelength $λ_{e}$ . The longest wavelength of radiation that can emit photoelectron is $λ_{0}$ . Expression for de-Broglie wavelength is given by:

(m : mass of the electron, h : Planck's constant and c : speed of light) [2025]

1 $λ_{e} = \sqrt{\frac{h}{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

2 $λ_{e} = \sqrt{\frac{h λ_{0}}{2 m c}}$

3 $λ_{e} = \frac{h}{\sqrt{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

4 $λ_{e} = \sqrt{\frac{h λ_{i}}{2 m c}}$

Ans.
(1)

$E = ϕ + K E \Rightarrow \frac{h c}{λ_{e}} = ϕ + K E$

$λ_{e} = \frac{h}{\sqrt{2 m K . E}}, E = \frac{h c}{λ_{i}} and ϕ = \frac{h c}{λ_{0}}$

$\frac{h^{2}}{2 m λ_{e}^{2}} = \frac{h c}{λ_{i}} - \frac{h c}{λ_{0}} \Rightarrow λ_{e} = \sqrt{\frac{h}{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

Solution

1 $λ_{e} = \sqrt{\frac{h}{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

2 $λ_{e} = \sqrt{\frac{h λ_{0}}{2 m c}}$

3 $λ_{e} = \frac{h}{\sqrt{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

4 $λ_{e} = \sqrt{\frac{h λ_{i}}{2 m c}}$

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Solution

1 λe=h2mc(1λi–1λ0)

2 λe=hλ02mc

3 λe=h2mc(1λi–1λ0)

4 λe=hλi2mc

Ans. (1) E=ϕ+KE ⇒ hcλe=ϕ+KE λe=h2mK.E, E=hcλi and ϕ=hcλ0 h22mλe2=hcλi–hcλ0 ⇒ λe=h2mc(1λi–1λ0)

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1 $λ_{e} = \sqrt{\frac{h}{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

2 $λ_{e} = \sqrt{\frac{h λ_{0}}{2 m c}}$

3 $λ_{e} = \frac{h}{\sqrt{2 m c (\frac{1}{λ_{i}} - \frac{1}{λ_{0}})}}$

4 $λ_{e} = \sqrt{\frac{h λ_{i}}{2 m c}}$