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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 −1-1 and Statement −2-2. 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 v>v0v>v_0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax⁡\mathrm{K}_{\max } and V0\mathrm{V}_0 respectively. If the frequency incident on the surface doubled, both the Kmax⁡\mathrm{K}_{\max } and V0\mathrm{V}_0 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

  1. AStatement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
  2. BStatement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
  3. CStatement-1 is false, Statement-2 is true.
  4. 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}' vv ' 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 }}maynotbeequalto2may not be equal to 2\Rightarrow \frac{\mathrm{V}_0^{\prime}}{\mathrm{V}_0}maynotequalto2may not equal to 2\mathrm{KE} \max =\mathrm{hu}-\mathrm{hv}_0\mathrm{V}=\frac{h \mathrm{v}}{\mathrm{e}}-\frac{h \mathrm{~h}_0}{\mathrm{e}}$
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