Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy. \[{{T}_{A}}\]expressed in eV and the de-Broglie wavelength\[{{\lambda }_{A}}\]. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is\[{{T}_{B}}=({{T}_{A}}-1.5eV)\]. If the de-Broglie wavelength of these photoelectrons is\[{{\lambda }_{B}}=2{{\lambda }_{A}},\]then: (1) the work function of A is 2.25 eV (2) the work function of B is 4.20 eV (3) \[{{T}_{B}}=2.7\text{ }eV\] (4) \[{{T}_{A}}=2.00\text{ }eV\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: D
Solution :
\[KE=E-W\] Therefore, \[{{T}_{A}}=4.25-{{W}_{A}}\] ...(i) \[{{T}_{B}}=({{T}_{A}}-1.50)=4.70-{{W}_{B}}\] ...(ii) Eqs. (i) and (ii).give \[{{W}_{B}}-{{W}_{A}}=1.95\,eV\] ...(iii) de-Broglie wavelength is given by \[\lambda =\frac{\lambda }{\sqrt{2}KM}\]or \[\lambda \propto \frac{1}{\sqrt{K}}\] (\[K=KE\]of electron) \[\therefore \] \[\frac{{{\lambda }_{B}}}{{{\lambda }_{A}}}=\sqrt{\frac{{{K}_{A}}}{{{K}_{B}}}}\] Or \[z=\sqrt{\frac{{{T}_{A}}}{{{T}_{A}}-1.5}}\] This gives, \[{{T}_{A}}=2\,eV\] From Eq. (i), \[{{W}_{A}}=4.25-{{T}_{A}}=2.25\,eV\] From Eq. (iii), \[{{W}_{B}}=1.95+{{W}_{A}}=4.20eV\] \[{{T}_{B}}=4.70-{{W}_{B}}=4.70-4.20=0.50\text{ }eV\]
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