The wavelength of the electron emitted by a metal sheet of work function 5 eV when photons from EMR of wavelength 62 nm strike the metal plate .

Options

A. `82.667 "Å"`
B. 3.16 nm
C. 0.316 nm
D. `826. 67 "Å"`

Answer

To find the wavelength of the electron emitted from a metal sheet with a work function of 5 eV when photons of wavelength 62 nm strike it, we can follow these steps:

### Step 1: Calculate the energy of the incoming photons
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) = Planck's constant = \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) = speed of light = \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) = wavelength of the photon = \( 62 \, \text{nm} = 62 \times 10^{-9} \, \text{m} \)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{62 \times 10^{-9} \, \text{m}} \]

Calculating this gives:
\[ E \approx 3.2 \times 10^{-18} \, \text{J} \]

### Step 2: Convert the energy from Joules to electron volts
To convert Joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \approx \frac{3.2 \times 10^{-18} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 20 \, \text{eV} \]

### Step 3: Calculate the kinetic energy of the emitted electron
The kinetic energy (KE) of the emitted electron can be found using the equation:
\[ KE = E - \phi \]
where \( \phi \) is the work function of the metal (5 eV):
\[ KE = 20 \, \text{eV} - 5 \, \text{eV} = 15 \, \text{eV} \]

### Step 4: Convert kinetic energy to Joules
Convert the kinetic energy back to Joules:
\[ KE \approx 15 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.4 \times 10^{-18} \, \text{J} \]

### Step 5: Calculate the wavelength of the emitted electron
The de Broglie wavelength of the electron can be calculated using the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy:
\[ p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)).

Substituting for \( p \):
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

Substituting the values:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{\sqrt{2 \cdot 9.1 \times 10^{-31} \, \text{kg} \cdot 2.4 \times 10^{-18} \, \text{J}}} \]

Calculating this gives:
\[ \lambda \approx 3.16 \times 10^{-10} \, \text{m} \]

### Step 6: Convert the wavelength to nanometers
To convert meters to nanometers:
\[ \lambda \approx 3.16 \times 10^{-10} \, \text{m} = 0.316 \, \text{nm} \]

### Final Answer
The wavelength of the emitted electron is approximately \( 0.316 \, \text{nm} \).

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