what is the difference in energy (in joules) between the n = 5 and n = 4 states of a hydrogen atom? report…
what is the difference in energy (in joules) between the n = 5 and n = 4 states of a hydrogen atom? report your answer in proper scientific notation to 3 sig figs by filling in the following blanks with units entered last.\n\nquestion 7\n1 pts\ndetermine the energy difference (in joules) between the n = 9 and the ground state of a hydrogen atom. report you answer in proper scientific notation to 3 sig figs by filling in the following blanks with units given last.\n\nquestion 8\n1 pts\nwhat is the wavelength of light (in nm) emitted from an excited li²⁺ ion in the n = 6 state that relaxes into the n = 3 state. report your answer to 1 decimal place and do not include units in your answer.
Answer
Question 1
Explanation:
Step1: Recall energy formula for hydrogen atom
The energy of an electron in a hydrogen - atom is given by $E_n=- \frac{13.6}{n^{2}}\text{ eV}$. First, convert eV to joules ($1\text{ eV}=1.602\times 10^{-19}\text{ J}$).
Step2: Calculate $E_5$ and $E_4$
$E_5=- \frac{13.6}{5^{2}}\text{ eV}=- \frac{13.6}{25}\text{ eV}=-0.544\text{ eV}$. Converting to joules, $E_5=-0.544\times1.602\times 10^{-19}\text{ J}=- 8.71488\times 10^{-20}\text{ J}$. $E_4=- \frac{13.6}{4^{2}}\text{ eV}=- \frac{13.6}{16}\text{ eV}=-0.85\text{ eV}$. Converting to joules, $E_4=-0.85\times1.602\times 10^{-19}\text{ J}=-1.3617\times 10^{-19}\text{ J}$.
Step3: Find energy difference
$\Delta E=E_5 - E_4=(-8.71488\times 10^{-20})-(-1.3617\times 10^{-19})=(1.3617\times 10^{-19}-8.71488\times 10^{-20})\text{ J}=4.90212\times 10^{-20}\text{ J}\approx4.90\times 10^{-20}\text{ J}$
Answer:
$4.90\times 10^{-20}$
Question 2
Explanation:
Step1: Recall energy formula for hydrogen atom
The energy of an electron in a hydrogen - atom is given by $E_n=- \frac{13.6}{n^{2}}\text{ eV}$. Convert eV to joules ($1\text{ eV}=1.602\times 10^{-19}\text{ J}$). The ground - state is $n = 1$, $E_1=-13.6\text{ eV}=-13.6\times1.602\times 10^{-19}\text{ J}=-2.17872\times 10^{-18}\text{ J}$.
Step2: Calculate $E_9$
$E_9=- \frac{13.6}{9^{2}}\text{ eV}=- \frac{13.6}{81}\text{ eV}\approx - 0.1679\text{ eV}$. Converting to joules, $E_9=-0.1679\times1.602\times 10^{-19}\text{ J}=-2.69976\times 10^{-20}\text{ J}$.
Step3: Find energy difference
$\Delta E=E_9 - E_1=(-2.69976\times 10^{-20})-(-2.17872\times 10^{-18})=(2.17872\times 10^{-18}-2.69976\times 10^{-20})\text{ J}=2.1517224\times 10^{-18}\text{ J}\approx2.15\times 10^{-18}\text{ J}$
Answer:
$2.15\times 10^{-18}$
Question 3
Explanation:
Step1: Use the Rydberg formula for hydrogen - like ions
For a hydrogen - like ion with atomic number $Z$, the Rydberg formula is $\frac{1}{\lambda}=R_HZ^{2}(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}})$, where $R_H = 1.097\times 10^{7}\text{ m}^{-1}$, $Z = 3$ for $Li^{2 +}$, $n_i = 6$ and $n_f = 3$. $\frac{1}{\lambda}=1.097\times 10^{7}\text{ m}^{-1}\times3^{2}(\frac{1}{3^{2}}-\frac{1}{6^{2}})$. First, calculate $\frac{1}{3^{2}}-\frac{1}{6^{2}}=\frac{1}{9}-\frac{1}{36}=\frac{4 - 1}{36}=\frac{3}{36}=\frac{1}{12}$. Then $\frac{1}{\lambda}=1.097\times 10^{7}\text{ m}^{-1}\times9\times\frac{1}{12}=1.097\times 10^{7}\text{ m}^{-1}\times\frac{3}{4}=8.2275\times 10^{6}\text{ m}^{-1}$.
Step2: Solve for $\lambda$
$\lambda=\frac{1}{8.2275\times 10^{6}\text{ m}^{-1}}=1.2154\times 10^{-7}\text{ m}$. Convert to nm ($1\text{ m}=10^{9}\text{ nm}$), so $\lambda = 121.54\text{ nm}\approx121.5\text{ nm}$
Answer:
$121.5$