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 metal A, the ejected photoelectrons have maximum kinetic energy, \[{{T}_{A}}\]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\[{{T}_{B}}=({{T}_{A}}-1.50)eV\]. If the de-Broglie wave- length 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)\[{{T}_{A}}=2.00\text{ }eV\] (4) \[{{T}_{B}}=2.75\text{ }eV\]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 :
Consider metal A incident energy = work function + maximum kinetic energy of photoelectrons. \[\therefore \] \[4.25={{W}_{A}}+{{T}_{A}}\] ...(i) Kinetic energy\[=\frac{{{p}^{2}}}{2m}\]where p = momentum \[\therefore \]\[{{T}_{A}}=\frac{p_{A}^{2}}{2m}=\frac{1}{2m}{{\left( \frac{h}{{{\lambda }_{A}}} \right)}^{2}}\]by de-Broglie equation \[\therefore \]\[{{T}_{A}}=\frac{1}{2m}{{\left( \frac{h}{{{\lambda }_{A}}} \right)}^{2}}\] ?..(ii) Consider metal B \[4.7={{W}_{B}}+({{T}_{A}}-1.5)\] ...(iii) \[{{T}_{B}}=\frac{1}{2m}{{\left( \frac{h}{{{\lambda }_{B}}} \right)}^{2}}\] ?.(iv) From Eqs. (iv) and (ii) we get \[\frac{{{T}_{B}}}{{{T}_{A}}}={{\left( \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}} \right)}^{2}}\] \[\frac{{{T}_{A}}-1.5}{{{T}_{A}}}={{\left( \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}} \right)}^{2}}\] \[[\because {{T}_{B}}={{T}_{A}}-15]\] \[\frac{{{T}_{A}}-1.5}{{{T}_{A}}}={{\left( \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}} \right)}^{2}}\] \[[\because {{\lambda }_{B}}=2{{\lambda }_{A}}]\] \[4{{T}_{A}}-6.0={{T}_{A}}\] Or \[3{{T}_{A}}=6\] \[{{T}_{A}}=2.00\,eV\] \[{{W}_{A}}=4.25\,-{{T}_{A}}\] \[=4.25-2=2.25\,eV\] From Eq. (iii) \[{{W}_{B}}=4.7-({{T}_{A}}-1.5)\] \[=4.7-2+1.5\] \[{{W}_{B}}=4.20\,eV\] Again \[{{T}_{B}}={{T}_{A}}-1.5=2-1.5=0.5\,eV\]
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