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## Energy and momentum of photoelectrons

Energy:

$E=hν=λhc $

Momentum:

$p=λh =cE $

$E=hν=λhc $

Momentum:

$p=λh =cE $

A particle having a de Broglie wavelength of 1.0 $A_{0}$ is associated with a momentum of (given $h=6.6×10_{−34}$ Js)

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Energy:

$E=hν=λhc $

Momentum:

$p=λh =cE $

$E=hν=λhc $

Momentum:

$p=λh =cE $

A particle having a de Broglie wavelength of 1.0 $A_{0}$ is associated with a momentum of (given $h=6.6×10_{−34}$ Js)

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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.

$K_{max}=E−φ$

When a metallic surface is illuminate with radiation of wavelength $λ$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2λ$, the stopping potential is $4V $. The threshold wavelength for the metallic surface is:

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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 $

Photons with energy $5eV$ are incident on a cathode $C$ in a photoelectric cell. The maximum energy of emitted photo electrons is $2eV$. When photons of energy $6eV$ are incident. on $C$, no photoelectrons will reach the anode $A$, if the stopping potential of $A$ relative to $C$ is :

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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 $

When a metallic surface is illuminate with radiation of wavelength $λ$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2λ$, the stopping potential is $4V $. The threshold wavelength for the metallic surface is:

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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.

The work function of the surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the :

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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,

$λhc −1eV=21 mv_{2}$

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

$v_{2}=1.10461×10_{12}$

$⇒v=1.05×10_{6}$

$⇒v=10_{6}$

A photosensitive metallic surface has work function, h $_{0}$. If photons of energy 2h $_{0}$ fall on this surface, the electrons come out with a maximum velocity of 4 10$_{6}$ m/s. When the photon energy is increased to 5h$_{0}$, then maximum velocity of photo electrons will be

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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)

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) :

$(m_{e}=9×10_{−31}kg)$