Question

What will be energy in \(\mathrm{eV}\) of photons of \(\lambda\) - rays having wave length \(0.1 \AA\) coming out of excited nucleus of radium? $\left\\{\mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{h}=6.625 \times 10^{-34}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(42.12 \times 10^{4}\) (B) \(12.42 \times 10^{4}\) (C) \(22.41 \times 10^{4}\) (D) \(24.21 \times 10^{4}\)

Short answer

The energy of the photon in electron volts (eV) can be found using the given values and the formula for the energy of a photon. After converting the wavelength to meters and calculating the energy in Joules, we then convert the energy to electron volts (eV). The calculated energy is $12.42 \times 10^4 eV$, which corresponds to option (B).

Step-by-step solution

Convert wavelength to meters

First, we need to convert the wavelength from angstroms (Å) to meters (m): 1 Å = 10^{-10} meters. Wavelength of the λ-ray, λ = 0.1 Å = 0.1 x 10^{-10} meters.

Calculate the energy of the photon in Joules

We know the formula to find the energy of a photon: \[E = \frac{hc}{λ}\] where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength. Now, plug in the given values: \(h = 6.625 \times 10^{-34} J s\), \(c = 3 \times 10^8 m/s\), and \(\lambda = 0.1 \times 10^{-10} m\). \[E = \frac{6.625 \times 10^{-34} Js \times 3 \times 10^8 m/s}{0.1 \times 10^{-10} m}\] Solve the equation to find the energy E in Joules.

Convert energy to electron volts

We found the energy E in Joules. Now, we need to convert it to electron volts (eV). We know that: 1 eV = \(1.6 \times 10^{-19} J\). Let's convert the energy E from Joules to electron volts. \[E(eV) = \frac{E(J)}{1.6 \times 10^{-19}}\] Calculate the energy E in electron volts (eV).

Compare the result with the given options

Now that we've calculated the energy in eV, we can compare it to the given answer choices. Determine which choice matches the calculated energy.