A) \[1.5\text{ }eV\]
B) \[4\text{ }eV\]
C) \[\text{3 }eV\]
D) \[\text{2 }eV\]
Correct Answer: B
Solution :
| [b] de-Broglie wavelength\[\left( \lambda \right)\], |
| \[mv=\frac{h}{\lambda }=p=\sqrt{2m(KE)}\] |
| \[\therefore \lambda =\frac{h}{\sqrt{2mKE}}\] |
| \[\therefore \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\sqrt{\frac{{{K}_{B}}}{{{K}_{A}}}}=\sqrt{\frac{{{T}_{A}}-1.5}{{{T}_{A}}}}\] (as given) |
| Also, \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{1}{2}\] |
| On solving \[{{T}_{A}}=2ev\] |
| \[\therefore {{K}_{B}}={{T}_{A}}-1.5=0.5ev\] |
| \[\therefore \] Work function of metal B is |
| \[{{\phi }_{B}}={{E}_{B}}-{{K}_{B}}\] |
| \[=4.5-0.5=4ev\] |
| For A, \[{{\phi }_{A}}={{E}_{A}}-{{T}_{A}}=2ev\] |
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