When an electron orbiting in hydrogen atom in its ground state jumps to higher excited state, the de-Broglie wavelength associated with it

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Aicharya · Accepted Answer

As $\mathrm{v} \propto \frac{1}{\mathrm{n}}$, we can also write $	herefore \quad \mathrm{mv} \propto \frac{1}{\mathrm{n}} \quad \Rightarrow \mathrm{p} \propto \frac{1}{\mathrm{n}}$ Also, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad \Rightarrow \lambda \propto \mathrm{n}$ $	herefore \quad$ Wavelength will increase.

When an electron orbiting in hydrogen atom in its ground state jumps to higher excited state, the de-Broglie…

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