Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.
The energy of the emitted photon is given by:

where
h is the Planck constant
f is the photon frequency
Substituting

, we find

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
Answer:
6.44 × 10^10 N/C
Explanation:
Electric field due to the ring on its axis is given by
E = K q r / (r^2 + x^2)^3/2
Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.
r = 0.25 m, x = 0.5 m, q = 5 C
K = 9 × 10^9 Nm^2/C^2
E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2
E = 6.44 × 10^10 N/C
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Answer:
second one is correct that is right
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316