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## Energy and momentum of photoelectrons

Energy:

$E=hν=λhc $

Momentum:

$p=λh =cE $

$E=hν=λhc $

Momentum:

$p=λh =cE $

Most asked questions

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Energy:

$E=hν=λhc $

Momentum:

$p=λh =cE $

$E=hν=λhc $

Momentum:

$p=λh =cE $

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Einstein's photoelectric equation :

$K_{max}=λhc −φ=hν−φ$

The kinetic energy of the photoelectron coming out may be anything between zero and $(E−φ)$ where $E=λhc $ is the energy supplied to the individual electrons.

$K_{max}=E−φ$

$K_{max}=λhc −φ=hν−φ$

The kinetic energy of the photoelectron coming out may be anything between zero and $(E−φ)$ where $E=λhc $ is the energy supplied to the individual electrons.

$K_{max}=E−φ$

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The smallest magnitude of the anode potential which just stops the photocurrent is called the stopping potential.

This potential should stop even the ost energitic photoelectron. Hence

Tha value of maximum kinetic energy should be equal to $eV_{0}$

We know $K_{m}ax=hν−ϕ=λhc −ϕ$

$V_{0}=λehc −eϕ $

This potential should stop even the ost energitic photoelectron. Hence

Tha value of maximum kinetic energy should be equal to $eV_{0}$

We know $K_{m}ax=hν−ϕ=λhc −ϕ$

$V_{0}=λehc −eϕ $

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Example: An electron of charge e and mass m is accelerated from rest by a potential difference V. Find the de-Broglie wavelength.

Solution:

$21 mv_{2}=eV$

$mv=2eVm $

So de-broglie wavelength

$λ=mvh $

$=2meV h $

Solution:

$21 mv_{2}=eV$

$mv=2eVm $

So de-broglie wavelength

$λ=mvh $

$=2meV h $

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The maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation, but is independent of its intensity.

For a frequency of $ν$ of incident radiation, lower than the threshold frequency $ν_{0}$ no photoelectric emission is possible even if the intensity is large.

For a frequency of $ν$ of incident radiation, lower than the threshold frequency $ν_{0}$ no photoelectric emission is possible even if the intensity is large.

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$K_{max}=λhc −φ=21 mv_{max}$

then $v_{max}=m2(λhc −φ) $

The work function of a metal surface is 1 eV. A light of wavelength 3000$A_{0}$ is incident on it. The maximum velocity of the photoelectrons is found as:

then $v_{max}=m2(λhc −φ) $

The work function of a metal surface is 1 eV. A light of wavelength 3000$A_{0}$ is incident on it. The maximum velocity of the photoelectrons is found as:

The maximum K.E with which electrons comes out,

$λhc −1eV=21 mv_{2}$

$⇒9.12×3.14×1.6×10_{−19}×10_{31} =v_{2}$

$v_{2}=1.10461×10_{12}$

$⇒v=1.05×10_{6}$

$⇒v=10_{6}$

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De Broglie's wavelength is the wavelength associated with a massive particle,hypothesized by De Broglie that explains Bohr's quantised orbits by bringing in the wave-particle duality. It is written as

$λ=mvh $ (de broglie wavelength)

$λ=mvh $ (de broglie wavelength)