Answer:
2.51 Angstroms
Explanation:
For a particle in a one dimensional box, the energy level, En, is given by the expression:
En = n²π² ħ² / 2ma²
where n is the energy level, ħ² is Planck constant divided into 2π, m is the mass of the electron ( 9.1 x 10⁻³¹ Kg ), and a is the length of the one dimensional box.
We can calculate the change in energy, ΔE, from n = 2 to n= 3 since we know the wavelength of the transition ( ΔE = h c/λ ) and then substitute this value for the expresion of the ΔE for a particle in a box and solve for the length a.
λ = 207 nm x 1 x 10⁻⁹ m/nm = 2.07 x 10⁻⁷ m ( SI units )
ΔE = 6.626 x 10⁻³⁴ J·s x 3 x 10⁸ m/s / 2.07 x 10⁻⁷ m
ΔE = 9.60 x 10⁻¹⁹ J
ΔE(2⇒3) = ( 3 - 2 ) x π² x ( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )
9.60 x 10⁻¹⁹ J = π² x( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )
⇒ a = 2.51 x 10⁻¹⁰ m
Converting to Angstroms:
a = 2.51 x 10⁻¹⁰ m x 1 x 10¹⁰ Angstrom / m = 2.51 Angstroms
Answer:
Question Answer
The half-life of cobalt-60 is 5.26 years. How many half-lives have passed in 10.52 years? 2
12.5% of a radioactive sample are left. How many half-lives have passed? 3
After 3 half-lives, how much of a 400 gram sample of radioactive uranium remains? 50g
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Answer:
The general formula for the carboxylic acids is C nH 2n+1COOH (where n is the number of carbon atoms in the molecule, minus 1).
Explanation:
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Answer:
The electron in xenon are dropping more energy levels than helium
