10. If an electron jumps from second orbit to first orbit in hydrogen atom, it emits photon of

When an electron jumps from a higher energy level (second orbit) to a lower energy level (first orbit) in a hydrogen atom, it emits a photon. This is due to the conservation of energy. The energy difference between the two orbits is released in the form of a photon.

The energy of the emitted photon can be calculated using the formula:

E = hf

where: E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 Joule second), and f is the frequency of the photon.

The frequency can be calculated using the Rydberg formula:

1/λ = RZ^2 (1/n1^2 - 1/n2^2)

where: λ is the wavelength of the photon, R is the Rydberg constant (1.097 x 10^7 per meter), Z is the atomic number (for hydrogen, Z=1), n1 is the lower energy level (for the first orbit, n1=1), and n2 is the higher energy level (for the second orbit, n2=2).

After calculating the wavelength using the Rydberg formula, you can find the frequency using the speed of light (c = 3.00 x 10^8 meters per second):

f = c/λ

Finally, you can find the energy of the photon using the E = hf formula.