calculate the wavelength (in nm) of light associated with the transition from n=1 to n=3 in the hydrogen…

Question 23

Explanation:

Step1: Use Rydberg formula

The Rydberg formula for hydrogen - like atoms is $\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, where $R_H = 1.097\times 10^{7}\ m^{-1}$, $n_1 = 1$, and $n_2=3$.

Step2: Substitute values

$\frac{1}{\lambda}=1.097\times 10^{7}\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=1.097\times 10^{7}\left(1 - \frac{1}{9}\right)=1.097\times 10^{7}\times\frac{8}{9}\approx9.751\times 10^{6}\ m^{-1}$.

Step3: Solve for $\lambda$

$\lambda=\frac{1}{9.751\times 10^{6}\ m^{-1}}\approx1.0255\times 10^{-7}\ m$. Convert to nm: $\lambda = 102.55\ nm$. Rounding to appropriate significant - figures, $\lambda = 102.6$.

Answer:

102.6

Question 24

Brief Explanations:

The angular momentum quantum number $l$ has different values for different orbitals. For an f - orbital, $l = 3$.

Answer:

3

Question 25

Explanation:

Step1: Determine atomic weights

Atomic weight of N: $A_N=14.01\ g/mol$, H: $A_H = 1.01\ g/mol$, P: $A_P=30.97\ g/mol$, O: $A_O = 16.00\ g/mol$.

Step2: Calculate molecular weight of $(NH_4)_3PO_4$

In $(NH_4)_3PO_4$, there are 3 N atoms, 12 H atoms, 1 P atom, and 4 O atoms. $M=(3\times14.01)+(12\times1.01)+30.97+(4\times16.00)$ $M = 42.03+12.12 + 30.97+64.00$. $M=149.12$.

Answer:

149.12