Question

Find the ratio of energies of photons produced due
to transition of an electron of hydrogen atom from
its (1) Second permitted energy level to first energy
level (2) The highest energy level to first
energy level​

Answer 1

Explanation:

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:

here is ur answer

Explanation:

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.

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