JAMB UTME - Physics - 2010 | Green Bridge CBT

Pergunta 1 Relatório

The ground stated energy for the hydrogen atom is 5.44 x 10-19J . If an electron drops from zero energy to ground state, calculate the frequency of the emitted radiation. [h = 6.6 x 10-34 Js]

5.0 x 1016 Hz

2.0 x 1016 Hz

5.0 x 1015 Hz

2.0 x 1015 Hz

No correct option

Detalhes da Resposta

When an electron drops from a higher energy level to a lower energy level, energy is released in the form of electromagnetic radiation (photons). The frequency of the radiation can be calculated using the formula: E = hf where E is the energy of the photon, h is Planck's constant, and f is the frequency of the radiation. In this question, the energy of the photon is equal to the difference between the initial energy level (zero) and the final energy level (5.44 x 10^-19 J): E = 5.44 x 10^-19 J Planck's constant is given as h = 6.6 x 10^-34 Js. Substituting these values into the formula, we get: 5.44 x 10^-19 J = (6.6 x 10^-34 Js) x f Solving for f, we get: f = (5.44 x 10^-19 J) / (6.6 x 10^-34 Js) f ≈ 8.24 x 10^14 Hz Therefore, the frequency of the emitted radiation is approximately 8.24 x 10^14 Hz, which is not one of the given options.

Ler a nota de aula em Structure Of The Atom (Nigeria Only) (WAEC)






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340	=	V2
√289	=	√400

∴ V2 =	340√400
	_√289