In the ground state of a hydrogen atom,an ele | vedclass

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Question

(A) Energy absorbed by the electron $= 1.5 	imes 2.18 	imes 10^{-18} \ J = 3.27 	imes 10^{-18} \ J$. Energy required to escape (ionization energy)

Vedclass · Accepted Answer

$\frac{h 	imes 10^{24}}{\sqrt{1.962}}$

Vedclass · Answer

(A) Energy absorbed by the electron $= 1.5 	imes 2.18 	imes 10^{-18} \ J = 3.27 	imes 10^{-18} \ J$. Energy required to escape (ionization energy) $= 2.18 	imes 10^{-18} \ J$. Kinetic energy $(KE)$ of the emitted electron $= 	ext{Energy absorbed} - 	ext{Energy required} = (3.27 - 2.18) 	imes 10^{-18} \ J = 1.09 	imes 10^{-18} \ J$. Using the de Broglie wavelength formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_e KE}}$. Substituting the values: $\lambda = \frac{h}{\sqrt{2 	imes (9 	imes 10^{-31} \ kg) 	imes (1.09 	imes 10^{-18} \ J)}}$. $\lambda = \frac{h}{\sqrt{19.62 	imes 10^{-49}}} = \frac{h}{\sqrt{1.962 	imes 10^{-48}}}$. $\lambda = \frac{h 	imes 10^{24}}{\sqrt{1.962}} \ m$.

In the ground state of a hydrogen atom,an electron absorbs $1.5$ times the minimum energy $\left(2.18 \times 10^{-18} \ J\right)$ required to escape from the atom. The wavelength of the emitted electron (in $m$) is $\left(m_e = 9 \times 10^{-31} \ kg\right)$.

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