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(C) The energy transferred to the mercury atom is equal to the kinetic energy gained by the electron,which is $E = eV = 4.9 \ eV$.This energy correspo

Vedclass · Accepted Answer

$2525$

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(C) The energy transferred to the mercury atom is equal to the kinetic energy gained by the electron,which is $E = eV = 4.9 \ eV$.This energy corresponds to the transition from the first excited state to the ground state.The relationship between energy and wavelength is given by $E = \frac{hc}{\lambda}$.Rearranging for wavelength: $\lambda = \frac{hc}{E}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $E = 4.9 	imes 1.6 	imes 10^{-19} \ J$.$\lambda = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{4.9 	imes 1.6 	imes 10^{-19}} \ m$.$\lambda \approx 2.536 	imes 10^{-7} \ m = 2536 \ \mathring{A}$.Given the options provided,the closest value is $2525 \ \mathring{A}$.

An electron passing through a potential difference of $4.9 \ V$ collides with a mercury atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state in $\mathring{A}$?

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