An electromagnetic wave of wavelength is incident on a photosensitive surface of negligible work function. If mass is of photoelectron emitted from the surface has de-Broglie wavelength , then [2021]

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Q.
An electromagnetic wave of wavelength $' λ'$ is incident on a photosensitive surface of negligible work function. If $' m'$ mass is of photoelectron emitted from the surface has de-Broglie wavelength $λ_{d}$ , then [2021]

1 $λ = (\frac{2 h}{m c}) λ_{d}^{2}$

2 $λ = (\frac{2 m}{h c}) λ_{d}^{2}$

3 $λ_{d} = (\frac{2 m c}{h}) λ^{2}$

4 $λ = (\frac{2 m c}{h}) λ_{d}^{2}$

Ans.
(4)

As work function is negligible, therefore Kinetic energy of emitted electron = Energy of incident photon

i.e., $\frac{1}{2} m v^{2} = h υ$

$\Rightarrow \frac{p^{2}}{2 m} = \frac{h c}{λ} \Rightarrow p = \sqrt{\frac{2 m h c}{λ}}$

de-Broglie wavelength of emitted electrons is

$λ_{d} = \frac{h}{p} = \frac{h}{\sqrt{\frac{2 m h c}{λ}}}$

$\Rightarrow λ_{d} = \sqrt{\frac{h λ}{2 m c}} \Rightarrow λ = (\frac{2 m c}{h}) λ_{d}^{2}$

Solution

Q.
An electromagnetic wave of wavelength $' λ'$ is incident on a photosensitive surface of negligible work function. If $' m'$ mass is of photoelectron emitted from the surface has de-Broglie wavelength $λ_{d}$ , then [2021]

1 $λ = (\frac{2 h}{m c}) λ_{d}^{2}$

2 $λ = (\frac{2 m}{h c}) λ_{d}^{2}$

3 $λ_{d} = (\frac{2 m c}{h}) λ^{2}$

4 $λ = (\frac{2 m c}{h}) λ_{d}^{2}$

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Solution

Q. An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then [2021]

1 λ=(2hmc)λd2

2 λ=(2mhc)λd2

3 λd=(2mch)λ2

4 λ=(2mch)λd2

Ans. (4) As work function is negligible, therefore Kinetic energy of emitted electron = Energy of incident photon i.e., 12mv2=hυ ⇒p22m=hcλ ⇒ p=2mhcλ de-Broglie wavelength of emitted electrons is λd=hp=h2mhcλ ⇒λd=hλ2mc ⇒ λ=(2mch)λd2

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Q.
An electromagnetic wave of wavelength $' λ'$ is incident on a photosensitive surface of negligible work function. If $' m'$ mass is of photoelectron emitted from the surface has de-Broglie wavelength $λ_{d}$ , then [2021]

1 $λ = (\frac{2 h}{m c}) λ_{d}^{2}$

2 $λ = (\frac{2 m}{h c}) λ_{d}^{2}$

3 $λ_{d} = (\frac{2 m c}{h}) λ^{2}$

4 $λ = (\frac{2 m c}{h}) λ_{d}^{2}$