Question

In a photoelectric experiment, a parallel beam of monochromatic light with a power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500V is applied between the cathode and anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experience a force F = n×10⁻⁴ N due to the impact of the electrons. The value of n is ______.

Mass of the electron mₑ=9×10⁻³¹ kg and 1.0 eV=1.6×10⁻¹⁹ J

Answer 1

The value of n can be determined using the formula for force: F = qE, where F is the force, q is the charge, and E is the electric field.

Step-by-step explanation:

The value of n can be determined using the formula for force:

F = qE

where F is the force, q is the charge, and E is the electric field. In this case, the force experienced by the anode is given as F = n×10⁻⁴ N. We can rearrange the formula to solve for n:

n = F / 10⁻⁴

Substituting the given value for F, the equation becomes:

n = (n×10⁻⁴) / 10⁻⁴

Therefore, the value of n is 1.