Question

1. what is the maximum number of electrons that can be placed in a p sublevel?

a. 2 b. 4 c. 6 d. 10
use 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.

1. calculate the energy in a light wave if the wavelength is 6.6 x 10^-7 m.

a. 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

1. 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?

a. 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

1. what is the energy of a photon of light with a frequency of 20000 hz?

a. 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

1. the energy of a photon of light was determined to be 4.5 x 10^-19 j. what is the wavelength of this light?

a. 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

Explanation:

16.

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 $2\times3 = 6$.

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 $E=\frac{6.63\times 10^{-34}\ J\cdot s\times3\times 10^{8}\ m/s}{6.6\times10^{-7}\ m}$
$E=\frac{19.89\times10^{-26}}{6.6\times10^{-7}}\ J=3\times 10^{-19}\ J$

Step1: Use the energy - frequency formula

The formula for the energy of a photon is $E = h
u$, where $E$ is energy, $h = 6.63\times10^{-34}\ J\cdot s$ and $
u$ is frequency.

Step2: Rearrange to solve for frequency

$
u=\frac{E}{h}$. Given $E = 7.59\times10^{-19}\ J$, then $
u=\frac{7.59\times10^{-19}\ J}{6.63\times10^{-34}\ J\cdot s}\approx1.14\times10^{15}\ Hz$

Answer:

c. 6

17.