Question

A metallic sphere of radius $r$ remote from all other bodies is irradiated with a radiation of wavelength $\lambda$ which is capable of causing photoelectric effect.

• A. the maximum potential gained by the sphere will be independent of its radius
• B. the net positive charge appearing on the sphere after a long time will depend on the radius of the sphere
• C. the maximum kinetic energy of the electrons emanating from the sphere will keep on declining with time
• D. the kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the sphere.

Answer 1

Answer: (A), (B), (C), (D)

All options are correct: A, B, C, and D.

Answer 2

The problem describes a metallic sphere, initially neutral and isolated, being irradiated by light capable of causing the photoelectric effect. As photoelectrons are emitted, the sphere accumulates a positive charge, leading to an increasing positive potential. This potential opposes further electron emission.

A) The maximum potential gained by the sphere will be independent of its radius.

As electrons are emitted, the sphere becomes positively charged. This positive charge creates an electric potential on the sphere. The emission of electrons continues until the potential energy required to escape the sphere due to its own positive charge ($eV_{max}$) equals the initial maximum kinetic energy ($KE_{max, initial}$) of the photoelectrons. At this point, no more electrons can escape, and the photoelectric current stops.

The initial maximum kinetic energy of the photoelectrons is given by Einstein's photoelectric equation:

$KE_{max, initial} = \frac{hc}{\lambda} - \phi$

where $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of incident radiation, and $\phi$ is the work function of the metal.

Thus, the maximum potential $V_{max}$ gained by the sphere is:

$eV_{max} = \frac{hc}{\lambda} - \phi$

$V_{max} = \frac{1}{e} \left( \frac{hc}{\lambda} - \phi \right)$

This expression for $V_{max}$ depends only on the properties of the incident light ($\lambda$) and the metal ($\phi$), and fundamental constants ($h, c, e$). It does not depend on the radius $r$ of the sphere. Therefore, statement (A) is correct.

B) The net positive charge appearing on the sphere after a long time will depend on the radius of the sphere.

The maximum potential $V_{max}$ achieved by the sphere is related to the maximum charge $Q_{max}$ accumulated on it by the formula for the potential of a charged sphere:

$V_{max} = \frac{kQ_{max}}{r}$

where $k = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

From this, the maximum charge $Q_{max}$ is:

$Q_{max} = \frac{r V_{max}}{k}$

Since $V_{max}$ is independent of $r$ (as shown in A), $Q_{max}$ is directly proportional to $r$. This means the net positive charge accumulated on the sphere after a long time will depend on its radius. Therefore, statement (B) is correct.

C) The maximum kinetic energy of the electrons emanating from the sphere will keep on declining with time.

Initially, when the sphere is neutral, the maximum kinetic energy of the emitted electrons is $KE_{max, initial} = \frac{hc}{\lambda} - \phi$.

As electrons are emitted, the sphere accumulates a positive charge, and its potential $V$ increases with time. When an electron is emitted from the sphere, it has to overcome this positive potential. The kinetic energy of an electron that successfully escapes the sphere (i.e., reaches a point far away from it) is given by:

$KE_{effective} = KE_{max, initial} - eV$

Since the potential $V$ of the sphere increases with time as more electrons are emitted, the effective kinetic energy $KE_{effective}$ of the escaping electrons will decrease with time. This process continues until $V$ reaches $V_{max}$, at which point $KE_{effective}$ becomes zero and electron emission stops. Therefore, statement (C) is correct.

D) The kinetic energy of the most energetic electrons emanating from the sphere initially will be independent of the radius of the sphere.

"Initially" refers to the moment when the sphere is neutral, or its potential is negligible ($V=0$). At this initial moment, the maximum kinetic energy of the emitted photoelectrons is given solely by Einstein's photoelectric equation:

$KE_{max, initial} = \frac{hc}{\lambda} - \phi$

This expression depends only on the wavelength of the incident light and the work function of the metal. It does not contain the radius $r$ of the sphere. Therefore, statement (D) is correct.

In summary:

• Maximum Potential (A): The sphere charges positively until its potential energy $eV_{max}$ equals the initial maximum kinetic energy of photoelectrons ($KE_{max, initial} = h\nu - \phi$). Thus, $V_{max} = (h\nu - \phi)/e$, which is independent of radius.
• Net Charge (B): The maximum charge $Q_{max}$ is related to $V_{max}$ by $Q_{max} = rV_{max}/k$. Since $V_{max}$ is independent of $r$, $Q_{max}$ is directly proportional to $r$.
• Kinetic Energy Decline (C): As the sphere charges positively, its potential $V$ increases. The effective kinetic energy of escaping electrons is $KE_{effective} = KE_{max, initial} - eV$. As $V$ increases with time, $KE_{effective}$ decreases.
• Initial Kinetic Energy (D): Initially, the sphere is neutral, so its potential is zero. The kinetic energy of the most energetic electrons is simply $KE_{max, initial} = h\nu - \phi$, which is independent of the sphere's radius.