An atom radiates 1.5*10^-19 of energy when an electron jumps from one energy level to another energy level.what is the wavelength of the emitted radiation (h=6.6*10^-34),speed of light in air=3*10^8ms^-1
To find the wavelength of the emitted radiation, we can use the formula for the energy of a photon, which is given by:
E = hc/λ
Where:
E = energy of the photon (1.5*10^-19 J)
h = Planck's constant (6.6*10^-34 J*s)
c = speed of light in air (3*10^8 m/s)
λ = wavelength of the radiation
Rearranging the formula to solve for λ, we get:
λ = hc / E
Substitute the values:
λ = (6.6*10^-34 J*s * 3*10^8 m/s) / 1.5*10^-19 J
λ = (1.98*10^-25 J*m) / 1.5*10^-19 J
λ = 1.32*10^-6 m
Therefore, the wavelength of the emitted radiation is 1.32*10^-6 m or 1320 nm.