Q: In the photoelectric effect, the stopping potential depends upon:

Stopping potential $V0$ is given by $eV0 = K{\max} = h\nu - \phi0$. Thus, it depends upon the frequency of the incident radiation and the nature of the metal surface (work function), but it is independent of the intensity of incident radiation.

Question 1

Which of the following particles has the shortest de Broglie wavelength for a given kinetic energy?

Accepted Answer

The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$, where $m$ is the mass and $K$ is the kinetic energy. For a given kinetic energy, wavelength is inversely proportional to the square root of the mass ($\lambda \propto \frac{1}{\sqrt{m}}$). Among the given options,

Question 2

The work function of a metal is $4.5 	ext{ eV}$. What is the threshold wavelength for this metal?

Accepted Answer

The threshold wavelength $\lambda0$ is related to the work function $\phi0$ by the formula $\lambda0 = \frac{hc}{\phi0}$. Substituting $hc \approx 12400 	ext{ eV}\cdot	ext{Å}$ and $\phi0 = 4.5 	ext{ eV}$, we get $\lambda0 = \frac{12400}{4.5} \approx 2755 	ext{ Å}$ or approximately $276 	ext{ nm

Question 3

Davisson and Germer experiment demonstrated:

Accepted Answer

The Davisson-Germer experiment proved the wave nature of electrons through electron diffraction from a nickel crystal, confirming de Broglie's hypothesis of matter waves.

Question 4

In the photoelectric effect, the stopping potential depends upon:

Accepted Answer

Stopping potential $V0$ is given by $eV0 = K{\max} = h
u - \phi0$. Thus, it depends upon the frequency of the incident radiation and the nature of the metal surface (work function), but it is independent of the intensity of incident radiation.

Question 5

The de Broglie wavelength of an electron accelerated through a potential difference of $100 	ext{ V}$ is approximately:

Accepted Answer

The formula for the de Broglie wavelength of an electron accelerated through potential $V$ is $\lambda = \frac{12.27}{\sqrt{V}} 	ext{ Å}$. For $V = 100 	ext{ V}$, $\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 	ext{ Å}$ or $0.123 	ext{ nm}$.

Dual Nature of Radiation and Matter