A) The work function of A is 2.25 eV
B) The work function of B is 4.20 eV
C) \[{{T}_{A}}=2.00\ eV\]
D) \[{{T}_{B}}=2.75\ eV\]
Correct Answer: B
Solution :
Kmax = E ? W0 \ TA = 4.25 ? (W0)A ...(i) TB =(TA? 1.5)= 4.70 ? (W0)B ...(ii) Equation (i) and (ii) gives (W0)B ? (W0)A = 1.95 eV De Broglie wave length \[\lambda =\frac{h}{\sqrt{2mK}}\]\[\Rightarrow \lambda \propto \frac{1}{\sqrt{K}}\] Þ \[\frac{{{\lambda }_{B}}}{{{\lambda }_{A}}}=\sqrt{\frac{{{K}_{A}}}{{{K}_{B}}}}\] \[\Rightarrow 2=\sqrt{\frac{{{T}_{A}}}{{{T}_{A}}-1.5}}\] Þ TA = 2eV From equation (i) and (iii) WA = 2.25 eV and WB = 4.20 eV.You need to login to perform this action.
You will be redirected in
3 sec