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 a maximum kinetic energy\[{{E}_{A}}\,eV\]and de-Broglie wavelength\[{{\lambda }_{A}}\]. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is\[{{E}_{B}}=({{E}_{A}}-1.50)eV\]. 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) \[{{E}_{A}}=2.0\,eV\] (4) \[{{E}_{B}}=2.75\,eV\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: A
Solution :
de-Broglie wavelength, \[\lambda =\frac{h}{\sqrt{2mE}}\] Then, \[{{\lambda }_{A}}=\frac{h}{\sqrt{2m{{E}_{A}}}}\] and \[{{\lambda }_{B}}=\frac{h}{\sqrt{2m{{E}_{B}}}}\] \[\therefore \] \[\frac{{{\lambda }_{B}}}{{{\lambda }_{A}}}=\sqrt{\frac{{{E}_{A}}}{{{E}_{B}}}}\] Given, \[{{E}_{B}}=({{E}_{A}}-1.5),{{\lambda }_{B}}=2{{\lambda }_{A}}\] \[2=\sqrt{\frac{{{E}_{A}}}{{{E}_{A}}-1.5}}\] or \[4=\frac{{{E}_{A}}}{{{E}_{A}}-1.5}\] or \[{{E}_{A}}=\frac{6}{3}=2\,eV\] \[\therefore \] \[{{E}_{B}}=({{E}_{A}}-1.5)\] \[=(2-1.5)\] \[=0.5\text{ }eV\] According to Einstein's photoelectric equation \[hv={{\phi }_{0}}+K{{E}_{\max }}\] or \[{{\phi }_{0}}=hv-K{{E}_{\max }}\] \[{{\phi }_{A}}=4.25\,eV-2\,eV=2.25\,eV\] \[{{\phi }_{B}}=4.70\,eV-0.50\,eV=4.20\,eV\] Hence, option is correct.
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