The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface , having work function `5.01 eV` , when ultraviolet light of `200 nm` falls on it , must be

Options

A. `2.4 V`
B. `- 1.2 V`
C. `- 2.4 V`
D. `1.2 V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy of the incident photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \) eV·s
- \( c \) (speed of light) = \( 3 \times 10^8 \) m/s
- \( \lambda \) (wavelength) = \( 200 \) nm = \( 200 \times 10^{-9} \) m

First, we convert the wavelength from nanometers to meters:
\[ \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \]

Now, substituting the values into the formula:
\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{200 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E = \frac{1.24071 \times 10^{-6} \, \text{eV·m}}{200 \times 10^{-9} \, \text{m}} = 6.20355 \, \text{eV} \]

### Step 2: Use the photoelectric equation
The photoelectric equation is given by:
\[ E = \phi + KE_{max} \]
where:
- \( E \) is the energy of the incident photon,
- \( \phi \) is the work function of the material,
- \( KE_{max} \) is the maximum kinetic energy of the emitted photoelectrons.

Rearranging the equation to find \( KE_{max} \):
\[ KE_{max} = E - \phi \]

Given that the work function \( \phi = 5.01 \, \text{eV} \):
\[ KE_{max} = 6.20355 \, \text{eV} - 5.01 \, \text{eV} = 1.19355 \, \text{eV} \]

### Step 3: Relate kinetic energy to stopping potential
The maximum kinetic energy of the emitted electrons is also related to the stopping potential \( V_0 \) by:
\[ KE_{max} = eV_0 \]
where \( e \) is the charge of an electron (1 eV = 1 volt for an electron).

Thus, we can write:
\[ V_0 = KE_{max} \]

Substituting the value of \( KE_{max} \):
\[ V_0 = 1.19355 \, \text{V} \]

### Step 4: Round the answer
Rounding \( V_0 \) to two decimal places gives:
\[ V_0 \approx 1.19 \, \text{V} \]

### Final Answer
The potential difference that must be applied to stop the fastest photoelectrons emitted by the nickel surface is approximately:
\[ V_0 \approx 1.19 \, \text{V} \]

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