To solve the problem, let's break it down into two parts as per the question:
### Part (a): Find the frequency of radiation
1. **Understanding the energy of the emitted photoelectrons**:
The maximum energy of the photoelectrons emitted when the frequency is \( f \) can be expressed using the photoelectric equation:
\[
E_{max} = hf - \phi
\]
where \( E_{max} \) is the maximum kinetic energy of the photoelectrons, \( h \) is Planck's constant, and \( \phi \) is the work function of the metal.
2. **Ionization energy of hydrogen**:
The problem states that these photoelectrons can just ionize hydrogen atoms in the ground state. The ionization energy of hydrogen from the ground state is approximately \( 13.6 \, \text{eV} \). Therefore, we can write:
\[
hf - \phi = 13.6 \, \text{eV} \quad (1)
\]
3. **Considering the second frequency \( \frac{5}{6}f \)**:
When the frequency is \( \frac{5}{6}f \), the energy of the incident photons is:
\[
E = h \left( \frac{5}{6} f \right) = \frac{5}{6} hf
\]
The photoelectrons emitted can excite hydrogen atoms, which then emit radiation of wavelength \( 1215 \, \text{Å} \).
4. **Finding the energy of the emitted radiation**:
The wavelength \( \lambda = 1215 \, \text{Å} = 1215 \times 10^{-10} \, \text{m} \). The energy of the emitted radiation can be calculated using:
\[
E = \frac{hc}{\lambda}
\]
where \( c \) is the speed of light. Substituting the values:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1215 \times 10^{-10} \, \text{m}} \approx 1.64 \times 10^{-19} \, \text{J} \approx 10.2 \, \text{eV}
\]
5. **Setting up the equation for the second frequency**:
The energy of the emitted photoelectrons when the frequency is \( \frac{5}{6}f \) is:
\[
E_{max}' = \frac{5}{6}hf - \phi
\]
These photoelectrons can excite the hydrogen atom, so:
\[
\frac{5}{6}hf - \phi = 10.2 \, \text{eV} \quad (2)
\]
6. **Solving equations (1) and (2)**:
From equation (1):
\[
hf = \phi + 13.6
\]
Substituting \( hf \) in equation (2):
\[
\frac{5}{6}(\phi + 13.6) - \phi = 10.2
\]
Simplifying:
\[
\frac{5}{6}\phi + \frac{5}{6} \times 13.6 - \phi = 10.2
\]
\[
-\frac{1}{6}\phi + \frac{68}{6} = 10.2
\]
\[
-\frac{1}{6}\phi = 10.2 - 11.33
\]
\[
-\frac{1}{6}\phi = -1.13 \implies \phi = 6.78 \, \text{eV}
\]
7. **Finding the frequency \( f \)**:
Now substituting \( \phi \) back into equation (1):
\[
hf = 6.78 + 13.6 = 20.38 \, \text{eV}
\]
Using \( h = 4.1357 \times 10^{-15} \, \text{eV s} \):
\[
f = \frac{20.38}{4.1357 \times 10^{-15}} \approx 4.93 \times 10^{14} \, \text{Hz}
\]
### Final Answers:
(a) The frequency of radiation is approximately \( 4.93 \times 10^{14} \, \text{Hz} \).
(b) The work function of the metal is \( 6.78 \, \text{eV} \).
