What is wave length of light emitted when electron in hydrogen atom undergoes transition from an energy level n2=4 and n1=2
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HEY FRND HERE IS UR ANS.......
Since n2=4, n1=2
from formula :
E = 2.18×10^-18J[1/4^2 - 1/2^2]
E = 2.18×10^-18 × (-3/16)
E = -4.0875 × 10^18 J
Now,,
wavelength of light emitted,
lemda = hc / E
lemda = (6.626 × 10^-34)(3 × 10^8)/4.0875 × 10^-19
lemda = 4.8631 × 10^-7m
lemda = 486.3 × 10^-9m
lemda = 486 nm Ans... ;)
Hope it helps you✌✌
Thankyou ☺
SWEETY ⚡⚡
Since n2=4, n1=2
from formula :
E = 2.18×10^-18J[1/4^2 - 1/2^2]
E = 2.18×10^-18 × (-3/16)
E = -4.0875 × 10^18 J
Now,,
wavelength of light emitted,
lemda = hc / E
lemda = (6.626 × 10^-34)(3 × 10^8)/4.0875 × 10^-19
lemda = 4.8631 × 10^-7m
lemda = 486.3 × 10^-9m
lemda = 486 nm Ans... ;)
Hope it helps you✌✌
Thankyou ☺
SWEETY ⚡⚡
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