Question 509: Atoms & Nuclei

Question

When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes

Accepted Answer

De-Broglie wavelength $$\lambda \propto \frac{1}{p}$$ where $$\mathrm{p}$$ is momentum
For an electron velocity $$\mathrm{v} \propto \frac{1}{\mathrm{n}}$$,
hence $$\mathrm{p} \propto \frac{1}{\mathrm{n}}$$
$$	herefore \frac{\lambda_1}{\lambda_4}=\frac{1}{4}$$
$$	herefore \lambda_1=\frac{\lambda_4}{4}$$

Answer

$$\left(\frac{1}{2}ight)^{	ext {th }}$$

Answer

$$\left(\frac{1}{4}ight)^{	ext {th }}$$

Answer

$$\left(\frac{1}{8}ight)^{	ext {th }}$$

Answer

$$\left(\frac{1}{6}ight)^{	ext {th }}$$

Atoms & Nuclei

When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes

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