Question 497: Atoms & Nuclei

Question

The angular momentum of an electron moving in an orbit of hydrogen atom is $\mathrm{1.5\left(\frac{h}{\pi}\right)}$. The energy in the same orbit is nearly.

Accepted Answer

The given angular momentum of the electron is
$1.5\left(\frac{h}{\pi}ight)$
, where
$h$
is Planck's constant.
According to the Bohr model of the hydrogen atom, the allowed angular momenta for an electron are quantized and given by
$mvr = n \frac{h}{2\pi}$
where
$m$
is the mass of the electron,
$v$
is its velocity,
$r$
is the radius of the orbit, and
$n$
is the principal quantum number (an integer).
By comparing the given angular momentum with the quantized form, we get
$1.5 \frac{h}{\pi} = n \frac{h}{2\pi}$
, which simplifies to
$n = 3$
.
Now, we use the formula for the energy levels of the hydrogen atom:
$E_n = -\frac{13.6 	ext{eV}}{n^2}$
. Since
$n = 3$
, the energy can be calculated as
$E_3 = -\frac{13.6 	ext{eV}}{3^2}$
, which is approximately
$-1.5 	ext{eV}$
.

Answer

$-1.5$ eV

Answer

$-1.6$ eV

Answer

$-1.3$ eV

Answer

$-1.4$ eV

Atoms & Nuclei

The angular momentum of an electron moving in an orbit of hydrogen atom is \mathrm{1.5\left(\frac{h}{\pi}\right)}. The energy in the same orbit is nearly.

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