Question

plz calculate the ratio of energies of photons produced due to transition of electron of hydrogen atom from its a)second permitted energy level to the first level b) highest permitted energy level to the second permitted level.

Answer 1

For finding wavelength , we have to use the formula
1/λ = R[ 1/n₁² - 1/n₂² ]
where R = 1.0973 × 10⁷ m⁻¹ , n₁ is initial energy level and n₂ is final energy level

(a) n₁ = 1 and n₂ = 2
so, 1/λ = R[1/1² - 1/2² ] = 3R/4 = 3 × 1.0973 × 10⁷/4
λ = 1.215 × 10⁻⁷ m = 121.5 nm

Now, use Energy , E₁= 1240/121.5 = 10.2 eV

Again,
(b) n₁ = 2 and n₂ = ∞
1/λ = R[ 1/2² - 1/∞²] = R/4 = 1.0973 × 10⁷/4
λ = 4/1.0973 × 10⁻⁷m = 364.5 nm
Now, E₂= 1240/λ = 1240/364.5 = 3.4 eV

Now, ratio of E₁ and E₂ :
E₁ /E₂ = 10.20 eV/3.4eV = 1020/340 = 102/34 = 51/17 = 3/1

Answer 2

Answer:

amswer is 3/1

Explanation:

Three atomic states of a hydrogen like atom are shown in the figure. The transition from C to B yields a photon of wavelength 364.6 nm and the transition from B to A yields a photon of wavelength 121.5 nm, Then the transition from C to A will yield a photon of wavelength.