For the excited state of Ca at the absorption of 422.7 nm light,the energy difference is mathematically given as
E= 4.70x10-22 kJ/mol
<h3>What is the energy difference (kJ/mole) between the ground and the first excited state?</h3>
Generally, the equation for the Energy is mathematically given as
E = nhc / λ
Where
h= plank's constant
h= 6.625x 10-34 Js
c = speed of light
c= 3x 108 m/s
Therefore
E = 1*(6.625x 10-34 Js)( 3x 10^8 m/s) / ( 422.7x10^-9)
E= 4.70x10-22 kJ/mol
In conclusion, Energy
E= 4.70x10-22 kJ/mol
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Explanation: