To solve the problem, we need to find the threshold frequency of the photosensitive material when an electron in a hydrogen atom transitions from the first excited state to the ground state. Here's a step-by-step solution:
### Step 1: Determine the energy levels of the hydrogen atom
The energy levels of a hydrogen atom are given by the formula:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
where \( n \) is the principal quantum number.
### Step 2: Calculate the energy of the first excited state (n=2)
For the first excited state (n=2):
\[
E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV}
\]
### Step 3: Calculate the energy of the ground state (n=1)
For the ground state (n=1):
\[
E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
### Step 4: Calculate the energy difference (photon energy)
The energy difference when the electron jumps from n=2 to n=1 is:
\[
E = E_1 - E_2 = (-13.6 \, \text{eV}) - (-3.4 \, \text{eV}) = -13.6 + 3.4 = -10.2 \, \text{eV}
\]
Thus, the energy of the emitted photon is:
\[
E = 10.2 \, \text{eV}
\]
### Step 5: Relate stopping potential to photon energy
The stopping potential \( V_0 \) is given as 3.57 V. The energy of the photon can also be expressed in terms of the stopping potential:
\[
E = eV_0
\]
where \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \, \text{C} \)). Therefore:
\[
E = 3.57 \, \text{eV}
\]
### Step 6: Calculate the work function (Φ)
Using the relationship between photon energy, stopping potential, and work function:
\[
E = eV_0 + \Phi
\]
Substituting the values:
\[
10.2 \, \text{eV} = 3.57 \, \text{eV} + \Phi
\]
Solving for \( \Phi \):
\[
\Phi = 10.2 \, \text{eV} - 3.57 \, \text{eV} = 6.63 \, \text{eV}
\]
### Step 7: Calculate the threshold frequency (ν₀)
The work function is related to the threshold frequency by the equation:
\[
\Phi = h\nu_0
\]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)). We can convert the work function from eV to joules:
\[
\Phi = 6.63 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.0608 \times 10^{-18} \, \text{J}
\]
Now, we can solve for \( \nu_0 \):
\[
\nu_0 = \frac{\Phi}{h} = \frac{1.0608 \times 10^{-18} \, \text{J}}{6.63 \times 10^{-34} \, \text{Js}} \approx 1.6 \times 10^{15} \, \text{Hz}
\]
### Final Answer
The threshold frequency of the material is:
\[
\nu_0 \approx 1.6 \times 10^{15} \, \text{Hz}
\]
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