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 solve the problem of determining the minimum energy photon required to observe an object of size 2.5 Å (Angstroms), we can follow these steps:

### Step 1: Understand the relationship between energy, wavelength, and frequency
The energy (E) of a photon can be calculated using the formula:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the photon. We can also express this in terms of wavelength (\( \lambda \)):
\[ E = \frac{hc}{\lambda} \]
where \( c \) is the speed of light.

### Step 2: Convert the size of the object to the appropriate units
The size of the object is given as 2.5 Å. We need to keep this in Angstroms for our calculations since the constants we will use are also in Angstroms.

### Step 3: Use the known values for constants
Planck's constant \( h \) and the speed of light \( c \) can be combined into a single constant when using Angstroms:
\[ hc = 12400 \, \text{eV} \cdot \text{Å} \]

### Step 4: Substitute the values into the energy formula
Now we can substitute \( \lambda = 2.5 \, \text{Å} \) into the energy equation:
\[ E = \frac{hc}{\lambda} = \frac{12400 \, \text{eV} \cdot \text{Å}}{2.5 \, \text{Å}} \]

### Step 5: Calculate the energy
Perform the calculation:
\[ E = \frac{12400}{2.5} = 4960 \, \text{eV} \]

### Step 6: Convert the energy to kilo-electron volts
Since the answer options are in kilo-electron volts, we convert:
\[ E = 4.96 \, \text{keV} \]
which can be approximated to:
\[ E \approx 5 \, \text{keV} \]

### Step 7: Conclusion
The minimum energy photon that can be used to observe the object is approximately:
\[ \boxed{5 \, \text{keV}} \]

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