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Dual Nature of Radiation and Matter — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsDual Nature of Radiation and Matter4 Marks
Question
According to Einstein, when a photonoflight offrequencyu or wavelength $\lambda$ is incident on a photosensitive metal surface of work function $\phi_0$, where $\phi_0<\text{h}\upsilon \ ($here $, h$ is Planck's constant$),$ then the emission of photoelectrons takes place. The maximum kinetic energy of the emitted photoelectrons is given by $\text{K}_\text{max}=\text{h}\upsilon-\phi_0$. If the frequency of the incident light is $\upsilon_0$ called thresold frequency, the photoelectrons are emitted from metal without any kinetic energy. So $\text{h}\upsilon_0=\phi_0$.
  1. A metal of work function $3.3eV$ is illuminated by light of wavelength $300\ nm$. The maximum kinetic energy of photoelectrons emitted is $($taking $h = 6.6 \times 10^{-34} Js).$
  1. $0.413eV$
  2. $0.825eV$
  3. $1.65eV$
  4. $1.32eV$
  1. The variation of maximum kinetic energy $(K_{max})$ of the variation of maximum kinetic energy $(\upsilon)$ of the incident radiations can be represented by:
  1. The variation of photoelectric current $(i)$ with the intensity of the incident radiation $(I)$ can be represented by:
  1. The graph between the stopping potential $(V_0)$ and $\Big(\frac{1}{\lambda}\Big)$ is shown in the figure. $\phi_1,\phi_2,\phi_3$ are work function. Which of the following options is correct?
  1. $\phi_1:\phi_2:\phi_3=1:2:3$
  2. $\phi_1:\phi_2:\phi_3=4:2:1$
  3. $\phi_1:\phi_2:\phi_3=1:2:4$
  4. Ultraviolet tight can be used to emit photoelectrons from metal $2$ and metal $3$ only.
  1. Which of the following figures represent the variation of particle momentum and the associated de $-$ Broglie wavelength?
✓

Answer

  1. $(b) 0.825eV$
$\text{K}_\text{max}=\text{h}\upsilon-\phi_0=\frac{\text{hc}}{\lambda}-\phi_0$
$=\frac{(6.6\times10^{-34})\times(3\times10^8)}{(300\times10^{-9})\times(1.6\times10^{-19})}-3.3$
$=4.125\times3.3=0.825\text{eV} $
  1.  

$\text{K}_\text{max}=\text{h}\upsilon-\phi_0$, when $\upsilon=\upsilon_0,\text{K}_\text{max}=0$
$\therefore0=\text{h}\upsilon_0$ or $\phi_0=\text{h}\upsilon_0$
If $\upsilon<\upsilon_0$, then $K_\text{max}$ is negative, i.e., no photoelectric emission takes place.
Thus, graph $(c)$ is possible.​​​​​​​
  1.  

Photoelectric current $(i)$ is proportional to the intensity of the emission light.
Thus, graph $(a)$ is possible.
  1. (c) $\phi_1:\phi_2:\phi_3=1:2:4$
From Einstein's photoelectric equation,
$\text{K}_\text{max}=\text{eV}_0=\frac{\text{hc}}{\lambda}-\phi$
or $\text{V}_0=\frac{\text{hc}}{\text{e}}\cdot\frac{1}{\lambda}-\frac{\phi}{\text{e}}$
Graph of $V_0$ versus $\frac{1}{\lambda} s$ a straight line
Slope of straight line, $\tan\theta=\frac{\text{hc}}{\text{e}}$
At $V_0 = 0,$ we have
$\phi_1:\phi_2:\phi_3=\frac{\text{hc}}{\lambda_{01}}:\frac{\text{hc}}{\lambda_{02}}:\frac{\text{hc}}{\lambda_{03}}$
$0.001\text{hc}:0.002\text{hc}:0.004\text{hc}$
$\therefore1:2:4$

de $-$ Broglie wavelength
$\lambda=\frac{\text{h}}{\text{P}}$ i.e.,$\lambda\propto\frac{1}{\text{P}}$
So the graph between $\lambda$ and $p$ is of the type shown is option $(d)$.

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