16. what is the maximum number of electrons that can be placed in a p sublevel?\na. 2 b. 4 c. 6 d. 10\nuse…
16. what is the maximum number of electrons that can be placed in a p sublevel?\na. 2 b. 4 c. 6 d. 10\nuse the equations and constants from your formula chart to solve the following: show your work to earn partial credit. showing your work includes writing the equation, substituting, rearranging and solving.\n17. calculate the energy in a light wave if the wavelength is 6.6 x 10^-7 m.\na. 3 x 10^-19 j b. 6 x 10^-19 j c. 6.6 x 10^-7 j d. 3.3 x 10^-18 j\n18. light with 7.59 x10^-19 j of energy will eject an electron from a piece of silver. what is the frequency of this light?\na. 1.14 x10^15 hz b. 2.4 x 10^15 hz c. 3.7 x 10^26 hz d. 5.4 x 10^52 hz\n19. what is the energy of a photon of light with a frequency of 20000 hz?\na. 1.3 x 10^-19 j b. 1.3 x 10^-35 j c. 1.3 x 10^-29 j d. 1.3 x 10^-18 j\n20. the energy of a photon of light was determined to be 4.5 x 10^-19 j. what is the wavelength of this light?\na. 5.5 x 10^-19 m b. 4.42 x 10^-7 m c. 4.4 x 10^-67 m d. 5.0 x 10^-8 m
Answer
16.
Explanation:
Step1: Recall electron - capacity rule
Each orbital can hold a maximum of 2 electrons. A p - subshell has 3 orbitals ($p_x$, $p_y$, $p_z$).
Step2: Calculate number of electrons
The maximum number of electrons in a p - subshell is $2\times3 = 6$.
Answer:
c. 6
17.
Explanation:
Step1: Use the energy - wavelength formula
The formula for the energy of a photon is $E=\frac{hc}{\lambda}$, where $h = 6.63\times10^{-34}\ J\cdot s$ (Planck's constant), $c=3\times 10^{8}\ m/s$ (speed of light), and $\lambda$ is the wavelength.
Step2: Substitute the values
Given $\lambda = 6.6\times10^{-7}\ m$, then $E=\frac{6.63\times 10^{-34}\ J\cdot s\times3\times 10^{8}\ m/s}{6.6\times10^{-7}\ m}$ $E=\frac{19.89\times10^{-26}}{6.6\times10^{-7}}\ J=3\times 10^{-19}\ J$
Answer:
a. $3\times 10^{-19}\ J$
18.
Explanation:
Step1: Use the energy - frequency formula
The formula for the energy of a photon is $E = h\nu$, where $E$ is energy, $h = 6.63\times10^{-34}\ J\cdot s$ and $\nu$ is frequency.
Step2: Rearrange to solve for frequency
$\nu=\frac{E}{h}$. Given $E = 7.59\times10^{-19}\ J$, then $\nu=\frac{7.59\times10^{-19}\ J}{6.63\times10^{-34}\ J\cdot s}\approx1.14\times10^{15}\ Hz$
Answer:
a. $1.14\times10^{15}\ Hz$
19.
Explanation:
Step1: Use the energy - frequency formula
The formula is $E = h\nu$. Given $\nu=20000\ Hz = 2\times10^{4}\ Hz$ and $h = 6.63\times10^{-34}\ J\cdot s$.
Step2: Calculate the energy
$E=6.63\times10^{-34}\ J\cdot s\times2\times 10^{4}\ Hz = 1.326\times10^{-29}\ J\approx1.3\times10^{-29}\ J$
Answer:
c. $1.3\times10^{-29}\ J$
20.
Explanation:
Step1: First, use $E = h\nu$ and $c=\lambda\nu$ to get $\lambda=\frac{hc}{E}$
We know $h = 6.63\times10^{-34}\ J\cdot s$, $c = 3\times10^{8}\ m/s$ and $E = 4.5\times10^{-19}\ J$.
Step2: Substitute the values
$\lambda=\frac{6.63\times10^{-34}\ J\cdot s\times3\times10^{8}\ m/s}{4.5\times10^{-19}\ J}=\frac{19.89\times10^{-26}}{4.5\times10^{-19}}\ m=4.42\times10^{-7}\ m$
Answer:
b. $4.42\times10^{-7}\ m$