The radiation is emitted when a hydrogen atom | vedclass

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Question

(A) The energy difference $\Delta E$ is given by the formula $\Delta E = \frac{hc}{\lambda}$.Given: $h = 6.626 	imes 10^{-34} \, J \cdot s$,$c = 3

Vedclass · Accepted Answer

$3.0 	imes 10^{-19} \, J$

Vedclass · Answer

(A) The energy difference $\Delta E$ is given by the formula $\Delta E = \frac{hc}{\lambda}$.Given: $h = 6.626 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^8 \, m/s$,and $\lambda = 6.5 	imes 10^{-7} \, m$.Substituting the values:$\Delta E = \frac{(6.626 	imes 10^{-34} \, J \cdot s) 	imes (3 	imes 10^8 \, m/s)}{6.5 	imes 10^{-7} \, m}$$\Delta E \approx \frac{19.878 	imes 10^{-26}}{6.5 	imes 10^{-7}} \, J$$\Delta E \approx 3.058 	imes 10^{-19} \, J$.Rounding to the nearest provided option,the energy difference is $3.0 	imes 10^{-19} \, J$.

The radiation is emitted when a hydrogen atom goes from a high energy state to a lower energy state. The wavelength of one line in the visible region of the atomic spectrum of hydrogen is $6.5 \times 10^{-7} \, m$. The energy difference between the two states is:

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