Question

The graph between the maximum speed ($v_{max}$) of a photoelectron and frequency (v) of the incident radiation, in photoelectric effect is correctly represented by:

Answer 1

D

Answer 2

The maximum kinetic energy ($K_{max}$) of a photoelectron emitted from a metal surface when illuminated by light of frequency $\nu$ is given by Einstein's photoelectric equation:

$K_{max} = h\nu - W_0$

where $h$ is Planck's constant and $W_0$ is the work function of the metal.

The maximum kinetic energy is related to the maximum speed ($v_{max}$) by:

$K_{max} = \frac{1}{2} m v_{max}^2$

where $m$ is the mass of the electron.

Combining these equations, we get:

$\frac{1}{2} m v_{max}^2 = h\nu - W_0$

We are interested in the relationship between $v_{max}$ and $\nu$. Rearranging the equation to solve for $v_{max}$:

$v_{max}^2 = \frac{2}{m}(h\nu - W_0)$ $v_{max} = \sqrt{\frac{2h}{m}\nu - \frac{2W_0}{m}}$

For photoelectric emission to occur, the energy of the incident photon must be greater than or equal to the work function, i.e., $h\nu \ge W_0$. This implies $\nu \ge \frac{W_0}{h}$. The minimum frequency required for emission is the threshold frequency, $\nu_0 = \frac{W_0}{h}$. So, photoemission occurs only when $\nu \ge \nu_0$. For $\nu < \nu_0$, no photoelectrons are emitted, and thus $v_{max} = 0$.

When $\nu = \nu_0$, $h\nu_0 = W_0$, so $K_{max} = h\nu_0 - W_0 = 0$, which means $v_{max} = 0$. This indicates that the graph starts at the point $(\nu_0, 0)$ on the $(\nu, v_{max})$ plane.

For $\nu > \nu_0$, $h\nu - W_0 > 0$, so $v_{max} > 0$. As $\nu$ increases, $h\nu - W_0$ increases, and therefore $v_{max} = \sqrt{\frac{2}{m}(h\nu - W_0)}$ increases.

The relationship $v_{max} = \sqrt{\frac{2h}{m}\nu - \frac{2W_0}{m}}$ is not linear. It is of the form $y = \sqrt{ax - b}$, where $y = v_{max}$, $x = \nu$, $a = \frac{2h}{m}$, and $b = \frac{2W_0}{m}$. This represents the upper half of a parabola $y^2 = ax - b$ (or $y^2 = a(x - b/a)$) opening to the right, with its vertex at $(b/a, 0) = (\nu_0, 0)$.

Let's examine the curvature of the graph for $\nu \ge \nu_0$. Let $f(\nu) = \sqrt{\frac{2h}{m}\nu - \frac{2W_0}{m}}$. The first derivative is $\frac{dv_{max}}{d\nu} = \frac{1}{2} \left(\frac{2h}{m}\nu - \frac{2W_0}{m}\right)^{-1/2} \left(\frac{2h}{m}\right) = \frac{h/m}{\sqrt{\frac{2h}{m}\nu - \frac{2W_0}{m}}}$. For $\nu > \nu_0$, $\frac{dv_{max}}{d\nu} > 0$, so the graph is increasing.

The second derivative is $\frac{d^2v_{max}}{d\nu^2} = \frac{d}{d\nu} \left[ \frac{h}{m} \left(\frac{2h}{m}\nu - \frac{2W_0}{m}\right)^{-1/2} \right] = \frac{h}{m} \left(-\frac{1}{2}\right) \left(\frac{2h}{m}\nu - \frac{2W_0}{m}\right)^{-3/2} \left(\frac{2h}{m}\right) = -\frac{h^2}{m^2} \left(\frac{2h}{m}\nu - \frac{2W_0}{m}\right)^{-3/2}$. For $\nu > \nu_0$, $\left(\frac{2h}{m}\nu - \frac{2W_0}{m}\right)^{3/2} > 0$, so $\frac{d^2v_{max}}{d\nu^2} < 0$. A negative second derivative indicates that the curve is concave down.

Therefore, the graph of $v_{max}$ versus $\nu$ is a curve that starts from the point $(\nu_0, 0)$ on the $\nu$-axis and increases with $\nu$ in a concave down manner. For $\nu < \nu_0$, $v_{max} = 0$. This means the graph is a horizontal line segment along the $\nu$-axis from $\nu=0$ to $\nu=\nu_0$, and then a concave down curve starting from $(\nu_0, 0)$.

The question asks for the graph between $v_{max}$ and $\nu$. Based on the analysis, the correct graph should show $v_{max}=0$ for $\nu \le \nu_0$ and a concave down curve starting from $(\nu_0, 0)$ for $\nu > \nu_0$. Among the typical options provided for this question, the one that matches this description is the curve starting from a positive frequency intercept ($\nu_0$) and bending downwards (concave down) as frequency increases.