A. Define the following terms: i. Ground state, ii. Excited state, iii. The work function of certain metal is 3.6 x 10⁻¹⁹ J. If the metal is illuminated with light of wave length 3.5 x 10⁻⁷m, calculate: a. The stopping potential b. Cut off frequency

$$
f = \frac{100\pi}{2\pi}
$$
$$
f = 50 \text{ Hz}
$$
The frequency of the current is $\boxed{\text{50 Hz}}$.

• $\gamma$. Calculate the root mean square of the current.

$$
I_{rms} = \frac{2.5}{\sqrt{2}}
$$
$$
I_{rms} \approx \frac{2.5}{1.4142}
$$
$$
I_{rms} \approx 1.7677 \text{ A}
$$
The root mean square of the current is $\boxed{\text{1.77 A}}$ (to 3 significant figures).

11. c. Write down the relationship between refracting angle A of a triangle glass prism, the minimum deviation Dm and refractive index.
The relationship for the refractive index ($\mu$) of a prism at minimum deviation is given by:

where $A$ is the refracting angle of the prism and $D_m$ is the angle of minimum deviation.

11. d. Show that the pressure on a single molecule of a gas inside a box is given by $p = \frac{1}{3} \rho c^2$, where $\rho = \text{density}$ and $C = \text{speed}$.
This formula represents the pressure exerted by an ideal gas, not a single molecule. Here's the derivation for the pressure of an ideal gas:

When it collides elastically with a wall perpendicular to the x-axis, its momentum changes by $\Delta p_x = m(-v_x) - m(v_x) = -2mv_x$.
The time taken for the molecule to travel to the opposite wall and back is $\Delta t = \frac{2L}{v_x}$.
The force exerted by this molecule on the wall is $F_x = \frac{|\Delta p_x|}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$.

The pressure $P$ on the wall is $P = \frac{F}{A} = \frac{F}{L^2} = \frac{1}{L^3} \sum_{i=1}^{N} m v_{xi}^2$.
Since $L^3 = V$ (volume of the box), $P = \frac{m}{V} \sum_{i=1}^{N} v_{xi}^2$.

$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}$.
The mean square speed $c^2$ (or $\overline{v^2}$) is given by $c^2 = \overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2} = 3\overline{v_x^2}$.
So, $\overline{v_x^2} = \frac{c^2}{3}$.
Therefore, $\sum_{i=1}^{N} v_{xi}^2 = N \overline{v_x^2} = N \frac{c^2}{3}$.

This shows that the pressure exerted by an ideal gas is given by $\boxed{P = \frac{1}{3} \rho c^2}$.