JEE Main 2011
- Exam
- JEE Main
- Year
- 2011
- Subject
- Physics
- Chapter
- Dual Nature of Matter
JEE Main 2011 · Physics · Dual Nature of Matter
This question has Statement -1 and Statement -2 . Of the four...
This question has Statement and Statement . Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency (the threshold frequency). The maximum kinetic energy and the stopping potential are and respectively. If the frequency incident on the surface doubled, both the and are also doubled.
Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
Options
- AStatement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
- BStatement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
- CStatement-1 is false, Statement-2 is true.
- DStatement-1 is true, Statement-2 is false.
Correct answer
C. Statement-1 is false, Statement-2 is true.
Step-by-step solution
\begin{aligned}& \mathrm{KE}_{\max }=\mathrm{hu}-\mathrm{hu}}_{0} \\
& \mathrm{hu}-\mathrm{hu}=\mathrm{e} \times \Delta \mathrm{v}\\
& \mathrm{V}}_{0}=\frac{\mathrm{hu}}{\mathrm{e}}-\frac{\mathrm{hv_{0 }}}{\mathrm{e}}
\end{aligned}' ' is doubled
\begin{aligned}& \mathrm{KE}_{\max }=2 \mathrm{hu}-\mathrm{hu}}_{0} \\
& \mathrm{~V}}_{0}^{\prime}=(\Delta \mathrm{V})^{\prime}=\frac{2 \mathrm{hu}}{\mathrm{e}}-\frac{h \mathrm{u}_0}{\mathrm{e}}
\end{aligned}\frac{\mathrm{KE}_{\max }}{\mathrm{KE}_{\max }}\Rightarrow \frac{\mathrm{V}_0^{\prime}}{\mathrm{V}_0}\mathrm{KE} \max =\mathrm{hu}-\mathrm{hv}_0\mathrm{V}=\frac{h \mathrm{v}}{\mathrm{e}}-\frac{h \mathrm{~h}_0}{\mathrm{e}}$