Question

The transition energy of the electron from higher to lower level is $3.0\times10^{-19},J$. The wavelength of the emitted light is about ($h=6.625\times10^{-34},Js,\ c=3\times10^8,ms^{-1}$)

• A. $330,nm$
• B. $660,nm$
• C. $990,nm$
• D. $310,nm$
• E. $230,nm$

Hint:

Chemistry Tip: Visible red light lies near $650$--$700,nm$, so $660,nm$ is a quick realistic check.

Answer 1

The Correct Option is B

Concept:
Energy of emitted photon: $$E=\frac{hc}{\lambda}$$ Hence: $$\lambda=\frac{hc}{E}$$
Step 1: Substitute given values.
$$\lambda=\frac{(6.625\times10^{-34})(3\times10^8)}{3.0\times10^{-19}}$$
Step 2: Simplify powers.
$$\lambda=\frac{19.875\times10^{-26}}{3\times10^{-19}}$$ $$=6.625\times10^{-7}m$$
Step 3: Convert into nm.
$$1,nm=10^{-9}m$$ $$6.625\times10^{-7}m=662.5,nm$$ Approximately: $$\lambda \approx 660,nm$$
Hence correct option is (B). :contentReference[oaicite:1]{index=1}