An isolated atom of a certain element emits light of wavelength 513 nm when the

Question

An isolated atom of a certain element emits light of wavelength 513 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 414 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.

A quantum particle in an infinitely deep square well has a wave function that is given by 1(x) for 0 s x s L and is zero otherwise. (a) Determine the probability of finding the particle between x 0 and x L. P m (b) Use the result of this calculation and symmetry arguments to find the probability of finding the particle between x L and x- 2L. Do not re-evaluate the integral.

Explanation / Answer

let energy at sixt state is E6, energy at fifth state is E5 and energy at second state is E2.

as we know energy of emitted photon is equal to difference in energy of the states.

when the atom falls from fifth state to seconds state:

h*c/lambda1=E5-E2...(1)

where h=planck's constant

c=speed of light

lambda1=513 nm

when it falls from sixth excited state to second excited state:

h*c/lambda2=E6-E2...(2)

where lambda2=414 nm

subtracting equation 1 from equation 2:

h*c*((1/lambda2)-(1/lambda1))=E6-E5

==>E6-E5=h*c*(lambda1-lambda2)/(lambda1*lambda2)....(3)

if wavelength of the light radiated when atom makes transition from 6th to 5th state is lambda,

then h*c/lambda=E6-E5

from equation 3,

h*c/lambda=h*c*(lambda1-lambda2)/(lambda1*lambda2)

==>lambda=lambda1*lambda2/(lambda1-lambda2)

=513*414/(513-414)

=2145.3 nm

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