Question

Light photons each of energy 3.5 ×10^-19 J falls on a cathode of a photocell. The current through the cell is reduced to zero by taking the cathode to a potential +0.25 V relative to anode .The work function of the cathode is ?​

Answer 1

Ans. 3.1 * 10^-19

Explanation :

Answer 2

The work function of the cathode is $3.1\times10^-^1^9J$.

Explanation:

Given the energy of the incident light photon, $E=3.5\times10^-^1^9J$.

The stopping potential,  $V_0=0.25V$.

The formula for the work function of the surface is given as

$\phi=E-K_{max}$

Here, $K_{max}$ is the maximum kinetic energy of the phtoelectron can be determined from the stopping potential.

$K_{max}=eV_0$

Here, e is the charge on the electron.

$K_{max}=1.6\times10^-^1^9C\times0.25V$

$K_{max}=4\times10^-^2^0J$.

substitute the given values in above formula, we get

$\phi=3.5\times10^-^1^9J-0.4\times10^-^1^9J$

$\phi=3.1\times10^-^1^9J$

Thus, the work function of the cathode is $3.1\times10^-^1^9J$.

#Learn More: photoelectric effect.

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