A) \[1.8\]
B) \[2.5\]
C) \[5.6\]
D) \[1.4\]
Correct Answer: A
Solution :
From Einstein's photoelectric equation, \[\frac{hc}{{{\lambda }_{1}}}=\phi +\frac{1}{2}m{{\left( 2\text{v} \right)}^{2}}\] ...(i) and \[\frac{hc}{{{\lambda }_{2}}}=\phi +\frac{1}{2}m{{\text{v}}^{2}}\] ...(ii) As per question, maximum speed of photoelectrons in two cases differ by a factor 2 From eqn.(i) & (ii) \[\Rightarrow \,\frac{\frac{hc}{{{\lambda }_{1}}}-\phi }{\frac{hc}{{{\lambda }_{2}}}-\phi }=4\Rightarrow \frac{hc}{{{\lambda }_{1}}}-\phi =\frac{4hc}{{{\lambda }_{2}}}-4\phi \] \[\Rightarrow \,\frac{4hc}{{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{1}}}=3\phi \Rightarrow \phi =\frac{1}{3}hc\left( \frac{4}{{{\lambda }_{2}}}-\frac{1}{{{\lambda }_{1}}} \right)\] \[=\frac{1}{3}\times 1240\left( \frac{4\times 350-540}{350\times 540} \right)=1.8\,eV\]
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