An atom radiates 1.5 x \(10^{-19}\)J of energy when an electron jumps from one energy level to another. What is the wavelength of the emitted radiation? (Pl...

Answer Details

The formula for calculating the wavelength of emitted radiation is given by: \begin{equation*} \lambda =\frac{hc}{\Delta E} \end{equation*} where $\lambda$ is the wavelength, $h$ is Planck's constant, $c$ is the speed of light and $\Delta E$ is the energy difference between the two energy levels. Given that the energy radiated is 1.5 x $10^{-19}$ J and the other constants are: Planck’s constant = 6.6 x $10^{-34}$ Js and speed of light in air = 3 x $10^{8}$ ms$^{-1}$, we can substitute these values into the formula and solve for the wavelength: \begin{align*} \lambda &=\frac{hc}{\Delta E}\\ &=\frac{(6.6\times 10^{-34}\text{Js})(3\times 10^8\text{ms}^{-1})}{1.5\times 10^{-19}\text{J}}\\ &=1.32\times 10^{-6}\text{m} \end{align*} Therefore, the wavelength of the emitted radiation is 1.32 x $10^{-6}$ m. Thus, the correct option is (a) 1.32 x $10^{-6}$ m.