Five volt of stopping potential is needed for the photoelectrons emitted out of a surface of work function 2.2 eV by the radiation of wavelength

Options

A. `1719A`
B. `3444A`
C. `861A`
D. `3000A`

Explanation

To solve the problem, we need to determine the wavelength of the radiation that causes photoelectrons to be emitted from a surface with a given work function and stopping potential. Here are the steps to find the solution:

### Step 1: Understand the relationship between stopping potential, work function, and kinetic energy.
The stopping potential (V) is related to the maximum kinetic energy (KE_max) of the emitted photoelectrons. The equation is given by:
\[ KE_{max} = eV \]
where \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs).

### Step 2: Calculate the maximum kinetic energy of the photoelectrons.
Given that the stopping potential \( V = 5 \, \text{V} \):
\[ KE_{max} = eV = 1.6 \times 10^{-19} \, \text{C} \times 5 \, \text{V} = 8.0 \times 10^{-19} \, \text{J} \]

### Step 3: Convert the work function from eV to joules.
The work function \( W \) is given as \( 2.2 \, \text{eV} \). To convert this to joules:
\[ W = 2.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.52 \times 10^{-19} \, \text{J} \]

### Step 4: Apply the photoelectric equation.
The total energy of the incoming photons (E) is given by:
\[ E = W + KE_{max} \]
Substituting the values we have:
\[ E = 3.52 \times 10^{-19} \, \text{J} + 8.0 \times 10^{-19} \, \text{J} = 11.52 \times 10^{-19} \, \text{J} \]

### Step 5: Relate the energy of the photons to wavelength.
The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)):
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step 6: Solve for the wavelength.
Rearranging the equation gives:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{11.52 \times 10^{-19} \, \text{J}} \]

Calculating this:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{11.52 \times 10^{-19}} \approx 1.724 \times 10^{-7} \, \text{m} \]
Converting to angstroms (1 m = \( 10^{10} \) angstroms):
\[ \lambda \approx 1724 \, \text{Å} \]

### Final Answer:
The wavelength of the radiation is approximately \( 1724 \, \text{Å} \).

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