The frequency of radiation emitted when the electron falls
from n = 4 to n = 1 in a hydrogen atom will be (Given : ionization
energy of H=2.18 ×10⁻¹⁸J atom⁻¹ and h = 6.625 × 10⁻³⁴ J s )
(a) 1.54´1015 s⁻¹ (b) 1.03´1015 s⁻¹
(c) 3.08´1015 s⁻¹ (d) 2.00´1015 s⁻¹
Answers
answer : option (c) 3.08 × 10^15 s-¹
electron falls from n = 4 to n = 1 in hydrogen atom.
from Rydberg's wavelength equation,
1/λ = RZ²(1/n1² - 1/n2²)
⇒hc/λ = RhcZ²(1/n1² - 1/n2²)
we know E = hv = hc/λ
⇒E = hv = RhcZ²(1/n1² - 1/n2²)
⇒hv = 2.18 × 10^-18 Z² [1/n1² - 1/n2²] [value of Rhc = 2.18 × 10^-18 J/atom ]
⇒6.625 × 10^-34 × v = 2.18 × 10^-18 × (1)² [1/1² - 1/4²]
⇒6.625 × 10^-34 × v = 2.18 × 10^-18 × 15/16
⇒v = (2.18 × 10^-18 × 15)/(16 × 6.625 × 10^-34)
⇒v = 3.08 × 10^15 s^-1
hence option (c) is correct choice.
also read similar questions : By how much would the stopping potential for a given photosensitive surface go up if the frequency of the incident radia...
https://brainly.in/question/1119651
The frequency of radiation emitted when an electron falls from n=4 to n=1 in a H atom will be ( Given that: ionisation e...
https://brainly.in/question/363774