Question 299: Dual Nature of Matter and Radiation

Question

In case of photoelectric effect, the graph of measured stopping potential $\left(\mathrm{V}_0\right)$ against frequency ' $v$ ' of incident light is a straight line. The slope of this line multiplied by the charge of electron (e) gives

Accepted Answer

The slope of the graph of the stopping potential $$V_0$$ versus the frequency $$v$$ of the incident light in the photoelectric effect is given by the equation:
$$ V_0 = \frac{h}{e} v - \frac{\phi}{e} $$
where:
$$V_0$$ is the stopping potential,
$$h$$ is Planck's constant,
$$e$$ is the charge of the electron,
$$v$$ is the frequency of the incident light,
$$\phi$$ is the work function of the metal.
From this equation, it can be observed that the slope of the line is $$\frac{h}{e}$$. When this slope is multiplied by the charge of the electron ($$e$$), the product is the Planck's constant ($$h$$):
$$ \left( \frac{h}{e} \right) e = h $$
Therefore, the slope of the line multiplied by the charge of the electron gives Planck's constant.
Correct answer: Option B
- the Planck's constant.

Answer

the work function of the metal.

Answer

the Planck's constant.

Answer

the maximum kinetic energy of the ejected electrons.

Answer

the threshold frequency for photoejection from the metal.

Dual Nature of Matter and Radiation

In case of photoelectric effect, the graph of measured stopping potential \left(\mathrm{V}_0\right) against frequency ' v ' of incident light is a straight line. The slope of this line multiplied by the charge of electron (e) gives

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