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egoroff_w [7]
3 years ago
7

An electron confined in a one-dimensional box is observed, at different times, to have energies of 12 eV, 27 eV, and 48 eV. What

is the length of the box? 45. | The nucleus of a typical atom
Chemistry
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

l=3.5*10^{-10}m

Explanation:

From the question we are told that:

1st Energy     E_1=12eV=4(3eV)

2nd Energy  E_2=27eV=9(3eV)

3rd Energy   E_3=48eV=16(3eV)

 

Generally the equation for Energy E for electron in one dimensional box at ground state E_0 is mathematically given by

  E_0=\frac{h}{8ml^2}

  E_0=\frac{h}{8ml^2}

Therefore Length a is mathematically given as

l=\sqrt{\frac{h^2}{8mE_0} }

l=\sqrt{\frac{(6.625*10^{-34})^2}{8(9.1*19^{-31}{(3eV(1.6*10^{-19}))}}

l=3.5*10^{-10}m

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<u>For 1:</u> The product is phosphoric acid and the solution is acidic in nature.

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