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Question: When photons of energy \(4.25\;eV\) strike the surface of metal A, the ejected photoelectrons have m...

When photons of energy 4.25  eV4.25\;eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy TA{T_A} and de Broglie wavelength λA{\lambda _A}. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70  eV4.70\;eV is TB=(TA−1.50)  eV{T_B} = \left( {{T_A} - 1.50} \right)\;eV. If the de Broglie wavelength of these photoelectrons is λB=2λA{\lambda _B} = 2{\lambda _A}, then:
A. The work function of A is 2.25  eV2.25\;eV
B. The work function of B is 4.20  eV4.20\;eV
C. TA=2.00  eV{T_A} = 2.00\;eV
D. TB=2.75  eV{T_B} = 2.75\;eV

Explanation

Solution

When a photon hits a metal, the metal will eject photoelectron. This is called a photoelectric effect. The incident energy of the photon is equal to the algebraic sum of the kinetic energy of the ejected photon and the work function to eject the photoelectron from the metal. By using the given relation between the kinetic energy of the photoelectron in two cases, the correct answer can be calculated.

Formula used:
De Broglie wavelength, λ=hp\lambda = \dfrac{h}{p}
Where hh is the Planck’s constant and pp is the momentum of the particle.

The kinetic energy of the particle, T=12mv2T = \dfrac{1}{2}m{v^2}
Where mm is the mass of the particle and vv is the velocity of the particle.

Complete step by step solution:
Given: Energy of photon in case one, EA=4.25  eV{E_A} = 4.25\;eV
Energy of photon in case two, EB=4.70  eV{E_B} = 4.70\;eV
Given relations are TB=(TA−1.50)  eV{T_B} = \left( {{T_A} - 1.50} \right)\;eV and λB=2λA{\lambda _B} = 2{\lambda _A}.
In the metal A, the de Broglie wavelength, λA=hpA{\lambda _A} = \dfrac{h}{{{p_A}}}
By rearranging, we get
pA=hλA{p_A} = \dfrac{h}{{{\lambda _A}}}
The kinetic energy of the photoelectron emitted from metal A, TA=12mvA2{T_A} = \dfrac{1}{2}m{v_A}^2
Since, the momentum of moving particle is pA=mvA{p_A} = m{v_A}
Squaring on both sides, we get pA2=m2vA2  ⇒mvA2=pA2m{p_A}^2 = {m^2}{v_A}^2\; \Rightarrow m{v_A}^2 = \dfrac{{{p_A}^2}}{m}
Substituting the above value on the equation of TA{T_A}, we get
TA=pA22m{T_A} = \dfrac{{{p_A}^2}}{{2m}}
Substitute the value of pA{p_A} in above equation, we get
TA=h22m×λA2  ................................................(1){T_A} = \dfrac{{{h^2}}}{{2m \times {\lambda _A}^2}}\;................................................\left( 1 \right)
In the metal B, the de Broglie wavelength, λB=hpB{\lambda _B} = \dfrac{h}{{{p_B}}}
By rearranging, we get
pB=hλB{p_B} = \dfrac{h}{{{\lambda _B}}}
The kinetic energy of the photoelectron emitted from metal B, TB=12mvB2{T_B} = \dfrac{1}{2}m{v_B}^2
Since, the momentum of moving particle is pB=mvB{p_B} = m{v_B}
Squaring on both sides, we get pB2=m2vB2  ⇒mvB2=pB2m{p_B}^2 = {m^2}{v_B}^2\; \Rightarrow m{v_B}^2 = \dfrac{{{p_B}^2}}{m}
Substituting the above value on the equation of TB{T_B}, we get
TB=pB22m{T_B} = \dfrac{{{p_B}^2}}{{2m}}
Substitute the value of pB{p_B} in above equation, we get
TB=h22m×λB2  ................................................(2){T_B} = \dfrac{{{h^2}}}{{2m \times {\lambda _B}^2}}\;................................................\left( 2 \right)
By dividing equation (1) by (2), we get
TATB=(h22m×λA2)(h22m×λB2)\dfrac{{{T_A}}}{{{T_B}}} = \dfrac{{\left( {\dfrac{{{h^2}}}{{2m \times {\lambda _A}^2}}} \right)}}{{\left( {\dfrac{{{h^2}}}{{2m \times {\lambda _B}^2}}} \right)}}
Since the Planck’s constant and the mass of the photoelectron are the same in both cases.
TATB=(1λA2)(1λB2) TATB=λB2λA2  ............................................(3)  \dfrac{{{T_A}}}{{{T_B}}} = \dfrac{{\left( {\dfrac{1}{{{\lambda _A}^2}}} \right)}}{{\left( {\dfrac{1}{{{\lambda _B}^2}}} \right)}} \\\ \dfrac{{{T_A}}}{{{T_B}}} = \dfrac{{{\lambda _B}^2}}{{{\lambda _A}^2}}\;............................................\left( 3 \right) \\\
From the question, λB=2λA{\lambda _B} = 2{\lambda _A}. Thus, by rearranging we get λBλA=2\dfrac{{{\lambda _B}}}{{{\lambda _A}}} = 2
By substituting the above relation in equation (3), we get
TATB=22=4\dfrac{{{T_A}}}{{{T_B}}} = {2^2} = 4
Hence, by rearranging we get TA=4×TB{T_A} = 4 \times {T_B}
From the question, TB=(TA−1.50)  eV{T_B} = \left( {{T_A} - 1.50} \right)\;eV
Substitute the value of TA{T_A} in above relation, we get
TB=(4TB−1.50)  eV 1.5=4TB−TB  {T_B} = \left( {4{T_B} - 1.50} \right)\;eV \\\ 1.5 = 4{T_B} - {T_B} \\\
By rearranging, we get
3TB=1.5 TB=1.53 TB=0.5  eV  3{T_B} = 1.5 \\\ {T_B} = \dfrac{{1.5}}{3} \\\ {T_B} = 0.5\;eV \\\
Substitute the value of TB{T_B} in the equation of TA{T_A}, we get
TA=4×0.5  eV TA=2  eV  {T_A} = 4 \times 0.5\;eV \\\ {T_A} = 2\;eV \\\
By the law of conservation of energy in the first case, the energy of the photon is equal to the sum of the kinetic energy of the photoelectron emitted and the work function tends the photoelectron to emit.
Hence, EA=TA+WA{E_A} = {T_A} + {W_A}
Substitute the given values in the above equation, we get
4.25  eV=2  eV+WA WA=4.25−2 WA=2.25  eV  4.25\;eV = 2\;eV + {W_A} \\\ {W_A} = 4.25 - 2 \\\ {W_A} = 2.25\;eV \\\

By the law of conservation of energy in the second case, the energy of a photon is equal to the sum of the kinetic energy of the photoelectron emitted and the work function tends the photoelectron to emit.
Hence, EB=TB+WB{E_B} = {T_B} + {W_B}
Substitute the given values in the above equation, we get
4.70  eV=0.5  eV+WB WB=4.70−0.5 WB=4.2  eV  4.70\;eV = 0.5\;eV + {W_B} \\\ {W_B} = 4.70 - 0.5 \\\ {W_B} = 4.2\;eV \\\

∴\therefore Hence, the options (A), (B) and (C) are correct.

Note:
When a photon hits the metal surface, due to the photoelectric effect, the photoelectrons are emitted from the metal. It requires some energy of the incident photon to eject an electron. That energy is called the work function of the photoelectric effect. The work function will vary based on the attraction force between the nucleus and the valence electron.