exam 1- requires respondus lockdown browser\nfinal grade 64.5/100.5\nsubmitted 9/22/25, 1:52 pm\nreceipt…

exam 1- requires respondus lockdown browser\nfinal grade 64.5/100.5\nsubmitted 9/22/25, 1:52 pm\nreceipt: 4511aas201d4c8ea178ca2a1052809\nwhich of the following is the lowest mass?\na 2.000×10² cg\nb 2000 μg\nc 2.000×10⁻⁴ mg\nd 2.000×10⁻⁴ kg\n3 multiple choice\ncalculate 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)\na 4.27×10⁻²¹ j\nb 2.91×10⁻¹⁹ j\nc 1.55×10⁻¹⁹ j\nd 1.07×10⁻²¹ j
Answer
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).
Answer:
B. $2000\text{ }\mu\text{g}$
Explanation 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:
C. $1.55\times10^{-19}\text{ J}$