The first excited state of Ca is reached by absorption of 422.7 nm light. Find the energy difference (kJ/mole) between the groun

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2 years ago

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d and first excited state. Watch your units. A. 2.83 x 10-7 B. 4.70 x 10-22 C. 283 D. 4.70 x 10-19 E. 457

1 answer:

dedylja [7] · Answer 1 · 2023-03-29T21:32:55+03:00

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