Einstein Photoelectric Equation

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Einstein explained the various laws of photoelectric emission on the basis of Planck’s revolutionary idea of Planck’s Quantum theory(year 1900) “Energy emitted or absorbed is not continuous, but it is in the form of packets of energy called quanta. One such quantum of radiation is called a photon which travels with the speed of light“.

The photon is not a material particle. It is electrically neutral and its rest mass is zero.

Energy of each photon is given by E = hν

where h is called Plank’s constant and ν is the frequency of light.

Einstein assume that one electron is ejected from a metal surface if one photon of suitable frequency falls on it.

The energy of photon(=hν) is used in two ways :-

For liberating the electron from the metal surface(providing energy equals to work function Φ₀)
Rest of the energy(hν – Φ₀) is used for providing the maximum kinetic energy(K_max) to the emitted photoelectrons.

So we can write :-

hν = Φ₀ + K_max

Maximum K.E. of the photoelectron can be written as, K_max = ½m(v_max)²

$\displaystyle \therefore h\nu ={{\phi }_{0}}+\frac{1}{2}mv_{\max }^{2}$ …..(1)

$\displaystyle {{K}_{\max }}=\frac{1}{2}mv_{\max }^{2}=h\nu -{{\phi }_{0}}$ …..(2)

This is called Einstein’s Photoelectric equation(Einstein photoelectric equation).

$\displaystyle {{K}_{\max }}=h\nu -h{{\nu }_{0}}=h(\nu -{{\nu }_{0}})$ …..(3)

We can also write K_max= eV₀and ν = (c/λ) , ν₀ = (c/λ₀), so we get

$\displaystyle e{{V}_{0}}={{K}_{\max }}=h\left( \frac{c}{\lambda }-{{\frac{c}{\lambda }}_{0}} \right)=hc\left( \frac{1}{\lambda }-{{\frac{1}{\lambda }}_{0}} \right)\text{ }\!\!~\!\!\text{ }$

In case if the incident photo has threshold frequency(ν₀), then the energy of the photon is just sufficient to eject the electron from the metal surface without giving it any kinetic energy.

In that case, $\displaystyle h{{\nu }_{0}}={{\phi }_{0}}$ .

Explanation of laws of photoelectric effect from

Einstein photoelectric equation

As Einstein assumed that one electron is ejected from a metal surface if one photon of suitable frequency falls on it, therefore, number of electrons ejected per second from a metal surface depends upon the number of photons falling on the metal surface per second, i.e., on intensity of light. Hence if intensity of incident light is increased, the number of photons falling per second on the metal surface increases, which results in an increase in the number of photoelectrons ejected(First law of photoelectric emission).
If ν < ν₀, then from Einstein photoelectric equation (3), maximum K.E. is negative, which is not possible. So photoelectric emission is not possible below threshold frequency(Second law of photoelectric emission).
If ν > ν₀, then from Einstein photoelectric equation (3), maximum K.E. ∝ ν,⇒ maximum kinetic energy of the emitted photoelectrons depends upon the frequency of the incident radiation and is independent of the intensity of the incident radiation(Third law of photoelectric emission).
Einstein assumed that photoelectric effect is an elastic collision between a photon and an electron. So the energy of the incident photon is instantaneously transferred to the electron inside the metal. Hence there is no time lag between the incidence of photon and ejection of photoelectron(Fourth law of photoelectric emission).

Some numerical on Einstein photoelectric equation and Photoelectric Effect

Example 1.

When a beam of $10.6 eV$ photons of intensity 2Wm^-2 falls on a platinum surface of area $1.0 \times 10^{- 4} m^{2}$ and work function $5.6 eV,$ $53 %$ of the incident photon eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV).

Solution.

Intensity is given by

$\displaystyle I=\frac{Energy(E)}{Area(A)\times time(t)}=\frac{N(h\nu )}{A\times t}$

Where N is number of photons incident in t time.

$\displaystyle \Rightarrow \frac{N}{t}=\frac{IA}{hv}$

As $53 %$ of the incident photon eject photoelectrons, so

Number of photoelectrons emitted per sec $\displaystyle \frac{N}{t}=\frac{IA}{hv}\times \frac{53}{100}$

$\displaystyle \frac{N}{t}=\frac{2\times (1.0\times {{10}^{-4}})}{10.6\times 1.6\times {{10}^{-19}}}\times \frac{53}{100}=6.25\times {{10}^{11}}$

Minimum kinetic energy of photoelectrons, K_min = 0 and

Using Einstein photoelectric equation maximum kinetic energy of photoelectrons,

K_max = E – Φ₀ = (10.6 – 5.6)eV = 5 eV

Example 2.

A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength B = 2.5 mT at an angle of 60^∘ with the field direction, the maximum pitch of the helix described by the electron is found to be 2.7 mm. Find the work function of the metal in electron volt. Given that specific charge of electron is 1.76 x 10¹¹ C/kg.

Solution.

Pitch of helical path, $\displaystyle P=\left( v\cos \theta \right)T=\frac{vT}{2}$

(as θ = 60º)

Here $\displaystyle T=\frac{2\pi m}{qB}=\frac{2\pi }{B\alpha }$

$\displaystyle (\alpha =\frac{q}{m})$

Hence Pitch, $\displaystyle P=\frac{\pi v}{B\alpha }$

$\displaystyle \Rightarrow v=\frac{B\alpha P}{\pi }=\frac{(2.5\times {{10}^{-3}})\times (1.76\times {{10}^{11}})\times (2.7\times {{10}^{-3}})}{3.14}$

$\displaystyle \Rightarrow v=0.38\times {{10}^{6}}m{{s}^{-1}}$

Using Einstein photoelectric equation

E = Φ₀ + K_max

⇒ Φ₀ = E – K_max

$\displaystyle {{\phi }_{0}}=E-\frac{1}{2}m{{v}^{2}}$

$\displaystyle {{\phi }_{0}}=4.9-\frac{1}{2}\frac{(9.1\times {{10}^{-31}})\times {{(0.38\times {{10}^{6}})}^{2}}}{1.6\times {{10}^{-19}}}$

⇒ Φ₀ = (4.9 – 0.4) eV

⇒ Φ₀ = 4.5 eV

Example 3.

An ultraviolet light of wavelength 2000 angstrom causes photo emission from a surface. The stopping potential is 2 volt. Find the work function in electron volt and the maximum speed of the ejected photoelectrons.

Solution.

Using Einstein photoelectric equation,

$\displaystyle {{\phi }_{0}}=\frac{hc}{\lambda }-e{{V}_{0}}=\frac{(6.63\times {{10}^{-34}})\times (3\times {{10}^{8}})}{(2000\times {{10}^{-10}})\times (1.6\times {{10}^{-19}})}-2$

$\displaystyle \Rightarrow {{\phi }_{0}}=4.2eV$

Now, as $\displaystyle \frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}$

$\displaystyle \Rightarrow {{v}_{\max }}=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2(1.6\times {{10}^{-19}})(2)}{9.1\times {{10}^{-31}}}}$

⇒ v_max = 8.4 × 10⁵ ms^-1

Example 4.

Suppose the wavelength of the incident light in photoelectric effect experiment is increased from $300 0 Å$ to $304 0 Å$ . Find the corresponding change in the stopping potential.

Solution.

From Einstein photoelectric equation

$\displaystyle \Delta {{V}_{0}}=\frac{\Delta {{E}_{incident}}}{e}$

$\displaystyle \Rightarrow \Delta {{V}_{0}}=\frac{1}{e}\left[ \frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{2}}} \right]$

$\displaystyle \Rightarrow \Delta {{V}_{0}}=\frac{hc}{e}\left[ \frac{{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}{{\lambda }_{2}}} \right]=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.6\times {{10}^{-19}}}\left[ \frac{3040-3000}{3000\times 3040\times {{10}^{-10}}} \right]$

⇒ ΔV₀ = 0.0545 Volt

Example 5.

A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron as it comes out of the metal. Under the same assumptions find the maximum number of collisions the electron should suffer before it becomes unable to come out of the metal.

Solution.

Remember :-

$\displaystyle E(\text{in eV})=\frac{12375}{\lambda (in\text{ angstrom})}$

$\displaystyle \Rightarrow E=\frac{12375}{4000}=3.1eV$

Energy of electron after first collision

E₁ = (90 % of E) = 2.79 eV

Energy of electron after second collision

E₂ = (90 % of E₁) = 2.51 eV

Hence, kinetic energy of the electron as it comes out of out of the metal surface

KE = (2.51 – 2.2) eV = 0.31 eV

Now

Energy of electron after third collision

E₃ = (90 % of E₂) = 2.26 eV

Energy of electron after forth collision

E₄ = (90 % of E₃) = 2.03 eV < Φ₀

After forth collision, the energy of electron is less than the work function of the metal, so the electron will not be able to come out of the metal.

Example 6.

Monochromatic radiation emitted when electron on hydrogen atom jumps from first exited to the grounds state irradiates a photosensitive material. The stopping potential is measured to be 3.57V. The threshold frequency of the material is :

(A) 2.5 × 10¹⁵ Hz

(B) 4 × 10¹⁵ Hz

(D) 1.6 × 10¹⁵ Hz

Solution :-

For hydrogen atom, $E_{n}$ = – $13.6 / n^{2}$ eV
For ground state, $n = 1$
$∴$ $E_{1}$ = – $13.6 / 1^{2}$ = – $13.6$ $e V$
For first excited state, $n = 2$
${∴ E}_{2}$ = – $13.6 / 2^{2}$ = – $3.4$ $e V$
The energy of the emitted photon when an electron jumps from first excited state to ground state is
$h v$ = $E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV$
Now maximum kinetic energy,
$K_{m a x} = e V0 = 3.57 e V$
According to Einstein’s photoelectric equation
$K_{m a x} = hν - ϕ_{0}$
$ϕ_{0} = 10.2 e V - 3.57 e V = 6.63 e V$
Threshold frequency,

$\displaystyle {{\nu }_{0}}=\frac{{{\phi }_{0}}}{h}=\frac{6.63\times 1.6\times {{10}^{-19}}}{6.63\times {{10}^{-34}}}=1.6\times {{10}^{15}}Hz$

Hence option (D) is correct.

Next Topic :- Particle nature of light

Previous Topic :- Failure of wave theory to explain photoelectric effect

Comment (1)

Sir your notes are soo good glad to have such a perfect note for my study thanku sir

Einstein photoelectric equation | Einstein’s photoelectric Effect

Einstein photoelectric Equation

Explanation of laws of photoelectric effect from

Einstein photoelectric equation

Some numerical on Einstein photoelectric equation and Photoelectric Effect

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