Question

exam 1- requires respondus lockdown browser
final grade 64.5/100.5
submitted 9/22/25, 1:52 pm
receipt: 4511aas201d4c8ea178ca2a1052809
which of the following is the lowest mass?
a 2.000×10² cg
b 2000 μg
c 2.000×10⁻⁴ mg
d 2.000×10⁻⁴ kg
3 multiple choice
calculate the energy, in joules, required to excite a hydrogen - atom by causing an electronic transition from the n = 3 to the n = 5 principal energy level. recall that the energy levels of the h atom are given by eₙ=-2.18×10⁻¹⁸ j(1/n²)
a 4.27×10⁻²¹ j
b 2.91×10⁻¹⁹ j
c 1.55×10⁻¹⁹ j
d 1.07×10⁻²¹ j

Explanation:

Step1: Convert all mass - units to grams

Option A:

$2.000\times10^{2}\text{ cg}$
Since $1\text{ cg}=10^{- 2}\text{ g}$, then $2.000\times10^{2}\text{ cg}=2.000\times10^{2}\times10^{-2}\text{ g}=2\text{ g}$

Option B:

$2000\text{ }\mu\text{g}$
Since $1\text{ }\mu\text{g}=10^{-6}\text{ g}$, then $2000\text{ }\mu\text{g}=2000\times10^{-6}\text{ g}=2\times10^{-3}\text{ g}$

Option C:

$2.000\times10^{4}\text{ Mg}$
Since $1\text{ Mg}=10^{6}\text{ g}$, then $2.000\times10^{4}\text{ Mg}=2.000\times10^{4}\times10^{6}\text{ g}=2\times10^{10}\text{ g}$

Option D:

$2.000\times10^{-4}\text{ kg}$
Since $1\text{ kg}=10^{3}\text{ g}$, then $2.000\times10^{-4}\text{ kg}=2.000\times10^{-4}\times10^{3}\text{ g}=0.2\text{ g}$

Step2: Compare the values

We have the values: Option A is $2\text{ g}$, Option B is $2\times10^{-3}\text{ g}$, Option C is $2\times10^{10}\text{ g}$, Option D is $0.2\text{ g}$.
The smallest value is $2\times10^{-3}\text{ g}$ (Option B).

for the second - part:

Step1: Use the energy - level formula for hydrogen atom

The energy levels of a hydrogen atom are given by $E_{n}=- 2.18\times10^{-18}\text{ J}(\frac{1}{n^{2}})$.
The energy required for a transition from $n_{1} = 3$ to $n_{2}=5$ is $\Delta E=E_{5}-E_{3}$.
$E_{5}=-2.18\times10^{-18}\text{ J}(\frac{1}{5^{2}})=-2.18\times10^{-18}\text{ J}\times\frac{1}{25}=-8.72\times10^{-20}\text{ J}$
$E_{3}=-2.18\times10^{-18}\text{ J}(\frac{1}{3^{2}})=-2.18\times10^{-18}\text{ J}\times\frac{1}{9}\approx - 2.42\times10^{-19}\text{ J}$

Step2: Calculate the energy of the transition

$\Delta E=E_{5}-E_{3}=-8.72\times10^{-20}\text{ J}-(-2.42\times10^{-19}\text{ J})$
$\Delta E=-8.72\times10^{-20}\text{ J}+2.42\times10^{-19}\text{ J}$
$\Delta E=( - 8.72\times10^{-20}+24.2\times10^{-20})\text{ J}=1.55\times10^{-19}\text{ J}$

Answer:

B. $2000\text{ }\mu\text{g}$