We wish to observe an object which is `2.5 Å` in size. The minimum energy photon that can be used is

Options

A. 5 keV
B. 8 keV
C. 10 keV
D. 12 keV

Answer

To find the minimum energy photon that can be used to observe an object of size 2.5 Å (angstrom), we can use the relationship between energy and wavelength given by Planck's equation:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

### Step 1: Convert the wavelength from angstroms to meters
1 Å = \( 10^{-10} \) m, so:
\[ \lambda = 2.5 \, \text{Å} = 2.5 \times 10^{-10} \, \text{m} \]

### Step 2: Substitute the values into Planck's equation
Now, substitute the values of \( h \), \( c \), and \( \lambda \) into the equation:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{2.5 \times 10^{-10} \, \text{m}} \]

### Step 3: Calculate the energy
Calculating the numerator:
\[ 6.63 \times 10^{-34} \times 3 \times 10^{8} = 1.989 \times 10^{-25} \, \text{Jm} \]

Now, divide by the wavelength:
\[ E = \frac{1.989 \times 10^{-25}}{2.5 \times 10^{-10}} \]
\[ E = 7.956 \times 10^{-16} \, \text{J} \]

### Step 4: Convert energy from joules to electron volts
To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{7.956 \times 10^{-16}}{1.6 \times 10^{-19}} \approx 4960 \, \text{eV} \]

### Step 5: Convert to keV
Since \( 1 \, \text{keV} = 1000 \, \text{eV} \):
\[ E \approx 4.960 \, \text{keV} \]

### Conclusion
The minimum energy photon that can be used to observe an object of size 2.5 Å is approximately **5 keV**.

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