All Solutions
Page 857: Practice Problems
$E = 0.80 times 10^{-33}$ J
$textbf{Find:}$ frequency of oscillation, $f$
To compute for $f$, we use the formula in getting the quantum of energy such that:
$$
begin{align*}
E &= hf \
f &= dfrac{E}{h} \
&= dfrac{0.80 times 10^{-33}}{6.626 times 10^{-34}} \
&= boxed{1.21 text{ Hz}}
end{align*}
$$
$$
nu=frac{E}{h}=frac{0.80times10^{-33}text{ J}}{6.62times 10^{-34}text{ J s}}=1.2text{ Hz}.
$$
$n = 8.6 times 10^{32}$
$f = 1.5$ Hz
$textbf{Find:}$ Energy of oscillation, $E$
To compute for $E$, we use the formula in getting the quantized energy such that:
$$
begin{align*}
E &= nhf \
&= (8.6 times 10^{32}) times (6.626 times 10^{-34}) times (1.5) \
&= boxed{0.85 text{J}}
end{align*}
$$
$$
E=nhnu = 8.6times10^{32}times6.62times10^{-34}text{ J s}times 1.5text{ Hz} =0.85text{ J}.
$$
$E = 0.75$ J
$f = 0.5$ Hz
$textbf{Find:}$ quantum number of pendulum, $n$
To compute for $n$, we use the formula in getting the quantized energy such that:
$$
begin{align*}
E &= nhf \
n &= dfrac{E}{hf} \
&= dfrac{0.75 text{ J}}{(6.626 times 10^{-34} text{ Js})times(0.50 text{ Hz})} \
&= boxed{2.26 times 10^{33}}
end{align*}
$$
$$
E=nhnu.
$$
This gives for the quantum number $n$
$$
n=frac{E}{hnu} = frac{0.75text{ J}}{6.62times10^{-34}text{ J}times 0.5text{ Hz}} = 2.26times 10^{33}.
$$