Question 540: 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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