A photon and an electron both have wavelength ` 1 Å` .
The ratio of energy of photon to that of electron is

Options

A. 1
B. 0.012
C. 82.7
D. ` 10^(-10)`

Answer

To find the ratio of the energy of a photon to that of an electron, both having a wavelength of \(1 \, \text{Å}\) (angstrom), we can follow these steps:

### Step 1: Calculate the Energy of the Photon
The energy \(E_p\) of a photon can be calculated using the formula:
\[ E_p = \frac{hc}{\lambda} \]
Where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))
- \(\lambda\) is the wavelength of the photon (\(1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}\))

Substituting the values:
\[ E_p = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1 \times 10^{-10} \, \text{m}} \]
Calculating this gives:
\[ E_p = \frac{1.9878 \times 10^{-25} \, \text{Jm}}{1 \times 10^{-10} \, \text{m}} = 1.9878 \times 10^{-15} \, \text{J} \]

### Step 2: Calculate the Energy of the Electron
The energy \(E_e\) of an electron can be calculated using the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \quad \Rightarrow \quad p = \frac{h}{\lambda} \]
Where \(p\) is the momentum of the electron. The kinetic energy \(E_e\) of the electron can be expressed as:
\[ E_e = \frac{p^2}{2m} \]
Substituting for \(p\):
\[ E_e = \frac{1}{2m} \left(\frac{h}{\lambda}\right)^2 \]
Where \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)) and \(\lambda\) is the wavelength of the electron (\(1 \, \text{Å} = 1 \times 10^{-10} \, \text{m}\)).

Substituting the values:
\[ E_e = \frac{1}{2(9.11 \times 10^{-31})} \left(\frac{6.626 \times 10^{-34}}{1 \times 10^{-10}}\right)^2 \]
Calculating this gives:
\[ E_e = \frac{1}{2(9.11 \times 10^{-31})} \left(6.626 \times 10^{-24}\right)^2 \]
\[ E_e = \frac{1}{2(9.11 \times 10^{-31})} (4.39 \times 10^{-47}) \approx \frac{2.195 \times 10^{-47}}{1.822 \times 10^{-30}} \approx 1.204 \times 10^{-17} \, \text{J} \]

### Step 3: Calculate the Ratio of Energies
Now we can find the ratio of the energy of the photon to that of the electron:
\[ \frac{E_p}{E_e} = \frac{1.9878 \times 10^{-15}}{1.204 \times 10^{-17}} \approx 165.0 \]

### Conclusion
Thus, the ratio of the energy of the photon to that of the electron is approximately:
\[ \frac{E_p}{E_e} \approx 165.0 \]