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

Gujarat 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 300nm. The maximum kinetic energy of photoelectrons emitted is (taking h = 6.6 × 10-34 Js).
  1. 0.413eV
  2. 0.825eV
  3. 1.65eV
  4. 1.32eV

  1. The variation of maximum kinetic energy (Kmax) 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 (V0) 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

Explanation:

$\text{K}_\text{max}=\text{h}\upsilon-\phi_0=\frac{\text{hc}}{\lambda}-\phi_0$

$=\frac{(6.6\times10^{-34})\times(3\times10^8)}{(300\times10^{-9})\times(1.6\times10^{-19})}-3.3$

$=4.125\times3.3=0.825\text{eV} $

  1.  

Explanation:

$\text{K}_\text{max}=\text{h}\upsilon-\phi_0$, 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 Kmax is negative, i.e., no photoelectric emission takes place. Thus, graph (c) is possible.

  1.  

Explanation:

Photoelectric current (i) is proportional to the intensity of the emission light. Thus, graph (a) is possible.

  1. (c) $\phi_1:\phi_2:\phi_3=1:2:4$

Explanation:

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 V0 versus $\frac{1}{\lambda}$ s a straight line

Slope of straight line, $\tan\theta=\frac{\text{hc}}{\text{e}}$

At V0 = 0, we have

$\phi_1:\phi_2:\phi_3=\frac{\text{hc}}{\lambda_{01}}:\frac{\text{hc}}{\lambda_{02}}:\frac{\text{hc}}{\lambda_{03}}$

$0.001\text{hc}:0.002\text{hc}:0.004\text{hc}$

$\therefore1:2:4$

  1.  

Explanation:

de-Broglie wavelength

$\lambda=\frac{\text{h}}{\text{P}}\text{i.e.,}\lambda\propto\frac{1}{\text{P}}$

So the graph between $\lambda$ and p is of the type shown is option (d).

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