Unit 15: Modern Physics
Weightage : 15–18%
The ground state of a certain type of atom has energy \(-E_0\). What is the wavelength of a photon with enough energy to ionize an atom in the ground state and give the ejected electron a kinetic energy of \(2E_0\)?
Correct Answer: \( \boxed{\mathrm{A}} \)
The photon must provide enough energy to both ionize the atom and supply the kinetic energy of the emitted electron.
The ionization energy is
\( E_{\mathrm{ion}}=E_0 \)
The emitted electron has kinetic energy
\( K=2E_0 \)
Therefore, the photon energy must be
\( E_{\gamma}=E_{\mathrm{ion}}+K=E_0+2E_0=3E_0 \)
Since the energy of a photon is
\( E_{\gamma}=\dfrac{hc}{\lambda} \)
solving for the wavelength gives
\( \lambda=\dfrac{hc}{3E_0} \)
Therefore, the correct answer is (A).
The graph above shows the maximum kinetic energy of electrons released in the photoelectric effect as a function of the frequency of the incident light. This graph is often used as evidence of the particle nature of light.
The graph supports which of the following statements?
Correct Answer: \( \boxed{\mathrm{D}} \)
According to Einstein’s photoelectric equation,
\( K_{\mathrm{max}} = hf – \Phi \)
where:
\( K_{\mathrm{max}} \) = maximum kinetic energy of the emitted electron
\( h \) = Planck’s constant
\( f \) = frequency of the incident light
\( \Phi \) = work function of the metal
The graph intersects the frequency axis at a nonzero value, indicating a threshold frequency. When \( f=f_0 \),
\( hf_0=\Phi \)
For frequencies below \( f_0 \), no electrons are emitted because the photon energy is insufficient to overcome the work function.
The straight-line graph also shows that the maximum kinetic energy increases linearly with frequency. Since photon energy is
\( E=hf \)
higher-frequency light consists of higher-energy photons, producing photoelectrons with greater kinetic energy.
Therefore, the correct answer is (D).
A beam of ultraviolet light shines on a metal plate, causing electrons to be ejected from the plate as shown in the figure. The velocity of the ejected electrons varies from nearly zero to a maximum of \(1.6\times10^{6}\,\mathrm{m/s}\). If the brightness of the beam is increased to twice the original amount, what will be the effect on the number of electrons leaving the metal plate and the maximum velocity of the electrons?
| Option | Number of Electrons Ejected | Maximum Velocity of Ejected Electrons |
|---|---|---|
| (A) | Increases | Increases |
| (B) | Increases | Remains the same |
| (C) | Remains the same | Increases |
| (D) | Remains the same | Remains the same |
Correct Answer: \( \boxed{\mathrm{B}} \)
The maximum kinetic energy of photoelectrons is given by Einstein’s photoelectric equation:
\(K_{\mathrm{max}}=hf-\Phi\)
where \(h\) is Planck’s constant, \(f\) is the frequency of the incident light, and \(\Phi\) is the work function of the metal.
Doubling the brightness (intensity) of the ultraviolet light increases the number of photons striking the metal each second, but it does not change the energy of each photon because the frequency remains constant.
As a result, more photons eject more electrons, so the number of emitted electrons increases.
Since the photon energy \(hf\) is unchanged, the maximum kinetic energy and therefore the maximum velocity of the emitted electrons remain unchanged.
The relationship between maximum kinetic energy and maximum speed is
\(K_{\mathrm{max}}=\dfrac{1}{2}mv_{\mathrm{max}}^{2}\)
Because \(K_{\mathrm{max}}\) remains constant, \(v_{\mathrm{max}}\) also remains constant.
Therefore, the correct answer is (B).
