Please give me the answers of the following questions from chapter structure of​ atom class 11 its urgent plz help!

Dear student,

16. Let the number of electrons present in ion, S³⁺ = x

$$\begin{gathered} Number of neutrons = x+ 30.4 \% of x = 1.304 x \\ Sinec the ion is tripositive, \\ Number of electrons in neutral atom = x+ 3 \\ Number of protons in neutral atom = x+3 \\ Mass number =Number of protons + Number of neutrons \\ 56 = x +3 + 1.304x \\ 2.304 x= 53 \\ x = 23 \\ Number of protons = atomic number = x \\ +3 = 2+ 3 =26 \\ Symbol of the ion { }^{56}F{e_{26}}^{3+} \end{gathered}$$
17. According to Millikan experiment,


q = ne
$$\begin{gathered} Here, q= -1.282 \times 10{ }^{-18}C \\ e= -1.60 \times {10}^{-19} C \\ n= \left(\frac{-1.282 \times {10}^{-18}C}{-1.60\times {10}^{-19 }C}\right)= 8.002 \cong 8 electrons \end{gathered}$$


18. 
     $$\begin{gathered} \left(a\right) radius = 2.6/2 =1.3 A^{\circ } = 130 pm \\ \left(b\right) Given length of the arrangement= 1.6 cm = 1.6 \times {10}^{-2}m \\ Diameter of zinc atom = 2.6 \times {10}^{-10 }m \\ Number of zinc atom = \frac{1.6\times {10}^{-2}m}{2.6 \times {10}^{-10} m} = 6.153 \times {10}^7 \end{gathered}$$



19. For an ion, 
$$\begin{gathered} \frac{1}{\lambda } = R_H{Z}^2\left(\frac{1}{{n_1}^2}- \frac{1}{n_{2}2}\right) \\ For H{e}^{+} ion, Z= 2 , n_2= 4, n_1=2 \end{gathered}$$
$$\begin{gathered} \frac{1}{\lambda } = R_H\times 4 \left(\frac{1}{2^2}- \frac{1}{4^2}\right) = 3{R_H}_{/4} \\ For hydrogen spectrum ,\frac{1}{\lambda } = 3R_{H/4} , Z=1 \\ \frac{1}{\lambda } = R_H \times 1 \left(\frac{1}{{n_1}^2}- \frac{1}{{n_2}^2}\right) \\ {R}_H \left(\frac{1}{{n_1}^2}- \frac{1}{{n_2}^2}\right) = 3R_{H/4} \\ \left(\frac{1}{{n_1}^2}- \frac{1}{{n_2}^2}\right) = \left(\frac{3}{4}\right) \\ Solving the above equation , we get n_1= 1 and n_2 = 2 \end{gathered}$$
Thus, the transition is from n=2 to n=1

20.
Using the de broglie equation,

$$\begin{gathered} \lambda = \frac{h}{mv} \\ For calculating the velocity, we use the energy expression, \\ E= \frac{1}{2} m{v}^2 \\ {v}^2 = 2\times E/m \\ Substituting the values of E and m , we get \\ {v}^2 = \frac{2\times 3\times {10}^{-25}}{9.1\times {10}^{-31}} \\ v= 8.12 \times {10}^2 \\ Substituting this value in the de Broglie eqution, we get \\ \lambda =\frac{h}{mv}= \frac{6.626 \times {10}^{-34}}{9.1\times {10}^{-31}\times 8.12 \times {10}^2} \\ = 8.972 \times {10}^{-7}m \\ Hence, the wavelength of electron will be 8.972\times {10}^{-7 }m \end{gathered}$$

Regards