Sankalp NEET Full Test-2 Question-14 Solution

Question 14: The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV , 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV , which of these photosensitive surfaces may emit photoelectrons?

(1) K only

(2) Na only

(3) Cs only

(4) Both Na and K

Answer: Option (3)

Explanation:

According to the photoelectric effect, photoelectrons are emitted only when the energy of the incident radiation is greater than or equal to the work function of the metal.

E \ge \phi

The given energy of incident radiation is

E = 2.20 \, \text{eV}

The work function of Caesium is

\phi_{\text{Cs}} = 2.14 \, \text{eV}

Since

2.20 \, \text{eV} > 2.14 \, \text{eV}

photoelectrons will be emitted from Caesium.

The work function of Potassium is

\phi_{\text{K}} = 2.30 \, \text{eV}

Since

2.20\,\text{eV}<2.30\,\text{eV}

photoelectrons will not be emitted from Potassium.

The work function of Sodium is

\phi_{\text{Na}} = 2.75 \, \text{eV}

Since

2.20\,\text{eV}<2.75\,\text{eV}

photoelectrons will not be emitted from Sodium.

Hence, only Caesium can emit photoelectrons.

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