Ahydrogen atom in the ground state absorbs a 12.75ev photon. immedietly after the absorption, the atom undergoes a quantum jump to the next- lowest energy level. what is the wavelength of the photon emmitted in this quantum jump? (answer- 1876nm)

1870nm

Explanation:

The energy of the hydrogen atom in the quantum state n is E = - (13.6 eV) / n^2. The ground state, when n = 1 and E = - 13.6 eV.

after absorption,the energy of atom is -13.6 + 12.75 = -0.85 eV.

quantum number for this energy state is

-0.85 = -13.6/n^2

n^2 = 13.6/0.85 = 16

n =4.

The next-lower energy orbit has energy

E = -13.6/3^2 = -1.511

Then transition emits a photon of energy -0.85-(-1.511) = 0.661 eV.

you would use

Landa = (planck's constant)(speed of light) / energy

λ = hc/E

or in your case, λ = (6.63 x 10-34 * 3 x 10^8) / (0.661 / 6.24 x 10^18)

to get λ = 1870 nm

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.