Ignoring sign. which transition is associated with the greatest energy change [Solved]

For a hydrogen-like atom, classify these electron transitions by whether they result in the absorption or emission of light. Ignoring sign, which transition is associated with the greatest energy change n = 3 to n = 5 n = 3 to n = 2 n = 2 to n = 1 n = 1 to n = 3 For a hydrogen-like atom, classify these electron transitions by whether or not they consequence within the absorption or emission of sunshine. Ignoring signal, which transition is related to the best power change n = 3 to n = 5 n = 3 to n = 2 n = 2 to n = 1 n = 1 to n = 3

1)
when theres transition from increased state to decrease
state, it entails emission
when theres transition from decrease state to increased
state, it entails absorption
So, Absorption are:
n=3 to n=5
n=1 to n=3
emission are:
n=2 to n=1
n=3 to n=2
2)
a)
Right here photon will probably be captured and itll excite the
atom
1/lambda = R* (1/nf^2 1/ni^2)
R is Rydberg fixed. R = 1.097*10^7
1/lambda = R* (1/nf^2 1/ni^2)
1/lambda = 1.097*10^7* (1/5^2 1/3^2)
lambda = 1.29*10^-6 m
lambda = 1290 nm
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(1.29*10^-6
m)
= 1.541*10^-19 J
b)
Right here photon will probably be emitted
1/lambda = R* (1/nf^2 1/ni^2)
R is Rydberg fixed. R = 1.097*10^7
1/lambda = R* (1/nf^2 1/ni^2)
1/lambda = 1.097*10^7* (1/2^2 1/3^2)
lambda = 6.563*10^-7 m
lambda = 656 nm
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(6.563*10^-7
m)
= 3.029*10^-19 J
Since that is emitted power, signal will probably be
damaging
So,
E = 3.029*10^-19 J
C)
Right here photon will probably be emitted
1/lambda = R* (1/nf^2 1/ni^2)
R is Rydberg fixed. R = 1.097*10^7
1/lambda = R* (1/nf^2 1/ni^2)
1/lambda = 1.097*10^7* (1/1^2 1/2^2)
lambda = 1.215*10^-7 m
lambda = 122 nm
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(1.215*10^-7
m)
= 1.635*10^-18 J
Since that is emitted power, signal will probably be
damaging
So,
E = 1.635*10^-18 J
D)
Right here photon will probably be captured and itll excite the
atom
1/lambda = R* (1/nf^2 1/ni^2)
R is Rydberg fixed. R = 1.097*10^7
1/lambda = R* (1/nf^2 1/ni^2)
1/lambda = 1.097*10^7* (1/3^2 1/1^2)
lambda = 1.032*10^-7 m
lambda = 103 nm
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(1.032*10^-7
m)
= 1.926*10^-18 J
Clearly power is largest for d
Reply: n=1 to n=3

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This post is last updated on hrtanswers.com at Date : 1st of September – 2022