When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital,a photon of approximate wavelength 629.1 nm is emitted.The energy difference between these 2p and 2s orbitals is A)3.16 $\times$10^-²⁸ J B)3.16 $\times$ 10^-¹⁹ J C)3.16 $\times$ 10^-¹⁷ J D)1.25 $\times$ 10^-³¹ J E)none of these

Question

Quizplus · Accepted Answer

The energy difference can be calculated using the formula E = hc/λ, where h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength (629.1 nm or 629.1 x 10^-9 m). Substituting these values gives an energy difference of approximately 3.16 x 10^-19 J.

When an Electron in a 2p Orbital of a Particular ×\times×

When an Electron in a 2p Orbital of a Particular $\times$