A monochromatic light of frequency is incident on a metal surface and ejects photoelectrons. The photoelectrons having maximum kinetic energy are just able to ionise a hypothetical atom in ground state. The energy levels of hypothetical one-electron atom' are given by `E_(n)=-(29)/(n^(2)) eV`, where n = 1,2,3.....When whole experiment is repeated with an incident light of frequency `v//2`, the photoelectrons with maximum kinetic energy are just able to ionise hydrogen atom in ground state. Calculate the work function of the metal 

Answer

To solve the problem, we need to analyze the two cases given in the question regarding the photoelectric effect and the energy levels of the hypothetical atom and hydrogen atom.

### Step 1: Understand the Energy Levels
The energy levels of the hypothetical one-electron atom are given by:
\[ E_n = -\frac{29}{n^2} \text{ eV} \]
For the ground state (n=1):
\[ E_1 = -29 \text{ eV} \]

### Step 2: Write the Energy Conservation Equation for Case 1
In the first case, when light of frequency \( \nu \) is incident on the metal surface, the energy of the incident photons is:
\[ E = h\nu \]
The maximum kinetic energy (KE) of the emitted photoelectrons is given by:
\[ KE = E - \phi \]
where \( \phi \) is the work function of the metal. Since the maximum kinetic energy of the photoelectrons is just enough to ionize the hypothetical atom, we have:
\[ KE = -E_1 = 29 \text{ eV} \]
Thus, we can write:
\[ h\nu - \phi = 29 \text{ eV} \]
This gives us our first equation:
\[ h\nu = \phi + 29 \text{ eV} \tag{1} \]

### Step 3: Write the Energy Conservation Equation for Case 2
In the second case, when light of frequency \( \frac{\nu}{2} \) is incident, the energy of the incident photons is:
\[ E = h\left(\frac{\nu}{2}\right) = \frac{h\nu}{2} \]
The maximum kinetic energy of the emitted photoelectrons is just enough to ionize the hydrogen atom in the ground state, which has energy:
\[ E_H = -13.6 \text{ eV} \]
Thus, we have:
\[ KE = -E_H = 13.6 \text{ eV} \]
So, we can write:
\[ \frac{h\nu}{2} - \phi = 13.6 \text{ eV} \]
This gives us our second equation:
\[ h\nu = 2(\phi + 13.6 \text{ eV}) \tag{2} \]

### Step 4: Equate the Two Expressions for \( h\nu \)
From equation (1):
\[ h\nu = \phi + 29 \text{ eV} \]
From equation (2):
\[ h\nu = 2(\phi + 13.6 \text{ eV}) \]

Setting these equal to each other:
\[ \phi + 29 = 2(\phi + 13.6) \]

### Step 5: Solve for \( \phi \)
Expanding the right side:
\[ \phi + 29 = 2\phi + 27.2 \]

Rearranging gives:
\[ 29 - 27.2 = 2\phi - \phi \]
\[ 1.8 = \phi \]

### Final Answer
Thus, the work function of the metal is:
\[ \phi = 1.8 \text{ eV} \]

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