Question

Calculate the de-broglie wavelength of the electron orbiting in the n=2 state of hydrogen atom

Answer 1

Hi there !
Here's the answer:

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¶¶¶ POINTS TO REMEMBER

¶¶ The Energy of an Electron in Bohr's orbit of Hydrogen atom
$E_{n} = -\frac{2 \pi^{2} me^{4}Z^{2}}{n^{2}h^{2}(4 \pi \epsilon_{0})^{2}}$

$E_{n} = - 13.6 \frac{Z^{2}}{n^{2}}\: eV$

where, n = Shell Number

• For Hydrogen Atom , Z = 1

$\implies E_{n} = \frac{13.6}{n^{2}}\: eV$

¶¶ Debroglie Wavelength
$\lambda = \frac{h}{\sqrt{2mE_{n}}}$

where, h = planck constant = $6.63×10^{-34}\: J-s$
m = mass of electron = $9.1× 10^{-31}\: kg$

¶¶ $1\: eV = 1.6 × 10^{-19}\: J$

¶¶ $1\: {\AA} = 1 \times 10^{-10}\: m$

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¶¶¶ SOLUTION:

Given,
Shell Number, n = 2

substitute in $E_{n}$

Energy ,$E_{n} = \frac{13.6}{4}\: eV$

$\implies E_{n} = 3.4\: eV$

$\implies E_{n} = 3.4×10^{-19}\: J$

Substitute m, h, $E_{n}$ in debroglie wavelength formula

$\implies \lambda = \frac{6.63×10^{-34}}{\sqrt{2×9.1×10^{-31}×3.4×1.6×10^{-19}}}$

$\implies \lambda = 6.663× 10^{-10}\: m$

$\implies \lambda = 6.663\: {\AA}$

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