$$T = 2\pi\sqrt{\frac{L}{g}}$$
Where T is the period of the pendulum in seconds, L is the length of the pendulum in meters, and g is the acceleration due to gravity in meters per second squared.
Since the length of the pendulum is the same on the earth and on the moon, we can use the period on the earth to find the length of the pendulum:
$$L = \frac{T^2g}{4\pi^2}$$
Substituting the given values, we get:
$$L = \frac{(1.35 \text{ s})^2 (9.8 \text{ m/s}^2)}{4\pi^2} = 1.43 \text{ m}$$
Now, we can use the formula above to find the period on the moon:
$$T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.43 \text{ m}}{1.62 \text{ m/s}^2}} = 2.73 \text{ s}$$
Therefore, the period of the pendulum on the surface of the moon is 2.73 seconds.
