Question

When ultraviolet radiation of a certain frequency falls on a potassium target, the photoelectrons released can be stopped completely by a retarding potential of $0.6$V. If the frequency of the radiation is increased by $10\%$, this stopping potential rises to $0.9$V. The work function of potassium is$?$

• A. $2.0$eV
• B. $2.4$eV
• C. $3.0$eV
• D. $2.8$eV

Answer 1

Answer: (B)

$2.4$eV

Stopping potential $e{V}_s=h\nu -\varphi$

where $V_s$ is the stopping potential, $\nu$ is the frequency of incident radiation and $\varphi$ is the work function of potassium.

$\therefore$ $0.6eV=h\nu -\varphi$ ......(1)

New frequency of radiation ${\nu }^{\prime }=1.1\nu$

New stopping potential $V_s^{\prime }=0.9V$

$\therefore$ $e{V}_s^{\prime }=h{\nu }^{\prime }-\varphi$

Or $0.9eV=1.1h\nu -\varphi$ .....(2)

Subtracting (1) from (1), we get $0.3eV=.1h\nu$

$⟹$ $h\nu =3eV$

Putting this in equation (1), we get $0.6eV=3eV-\varphi$

$⟹$ $\varphi =3-0.6=2.4eV$