Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

option 3 i guess

Explanation:

hope it helps u

Answer 2

For Lyman series $\lambda_{max}-\lambda_{min}$ = 340 A°. find the same for paschen series ?

solution : for minimum wavelength of Lyman's series, n₂ = ∞ , n₁ = 1

1/$\lambda_{min}$ = RZ

(1)²[1/1² - 1/∞²] = R [ using Rydberg's equation]

so, $\lambda_{min}=\frac{1}{R}$

for maximum wavelength of Lyman's series,

n₂ = 2 to n₁ = 1

1/$\lambda_{max}$ = R(1)²[1/1² - 1/2²] = 3R/4

so, $\lambda_{max}=\frac{4}{3R}$

now $\lambda_{max}-\lambda_{min}=-\frac{1}{R}+\frac{4}{3R}=\frac{1}{3R}$ = 340 A°

for paschen's series

for minimum wavelength,

1/$\lambda_{min}$ = R(1)²[1/3² - 1/∞²] = R/9

⇒$\lambda_{min}=\frac{9}{R}$

for maximum wavelength,

1/$\lambda_{max}$ =R(1)²[1/3² - 1/4²] = 7R/144

$\lambda_{max}$ = 144/7R

so, $\lambda_{max}-\lambda_{min}=\frac{7R}{144}-\frac{R}{9}=\frac{81}{7R}$

but 1/3R = 340 A°

so 1/R = 340 × 3 A°

81/7R = 81/7 × 340 × 3 ≈ 11802 A°

Therefore the difference in the maximum and minimum wavelength of paschen's series is 11802A° .