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...
Question 1 Report
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? (Planck’s constant =6.6 x \(10^{-34}\)Js. Speed of light in air=3 x \(10^{8}ms^{-1}\)
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.
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