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Grade: 11

                        

Hydrogen sample is in 1 st excited state. Photon of energy 8eV is used to excite the H-Sample. Find the De-Broglie wavelength of the electron.

2 years ago

Answers : (1)

Eshan
askIITians Faculty
2095 Points
							Dear student,

Ionization energy of H-atom in first excited state=\dfrac{13.6}{2^2}eV=3.4eV

Hence excess energy of photon imparted to electron=8eV-3.4eV=4.6eV=7.36\times 10^{-19}J

Hence momentum=p\sqrt{2mE}
De-broglie wavelength=\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2mE}}=5.72A{\circ}
2 years ago
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