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Calculate the frequency(hz) and energy(kj/mol) if light emitted when a hydrogen electron transitions from n=7 to n=4

Sagot :

DALAU

Answer:

frequency(hz) = ∆E =-0.917×10-¹⁸ J/H

energy(kj/mol) = v=1.384×1015 S​​​​​​-1​.

Explanation:

According to Rydberg eqn​​​​-

   ∆E= -Rh ( 1/nf2-1/ni2)

where ∆E- The energy change caused

   Rh-Rydberg constant

nf-Principal quantum number of the destination level

ni-Principle quantum number of the initial level

   Value of Rh=2.18×10-18J/H

   Here nf=4 and ni=7

   Therefore ∆E=-2.18×10-18(1/4²-1/7²)

∆E=-0.917×10-¹⁸ J/H

The -ve sign indicates the release of energy from H atom due to relaxation

Frequency  v=∆E/h

where h is the planck's constant = 6.626×10-34 J.S

Hence v=0.917×10-18/6.626×10-34

v=1.384×1015 S​​​​​​-1​.

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