SATHEE: Modern Physics 1 Question 6

Modern Physics 1 Question 6

5. Taking the wavelength of first Balmer line in hydrogen spectrum $(n = 3$ to $n = 2)$ as $660 n m$ , the wavelength of the $2^{nd}$ Balmer line $(n = 4$ to $n = 2)$ will be

(Main 2019, 9 April I)

(a) $889.2 n m ($ b)

(b) $388.9 n m$

(d) $488.9 n m$

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Solution:

Expression for the energy of the hydrogenic electron states for atoms of atomic number $Z$ is given by

$\begin{aligned} E & = h ν = \frac{Z^{2} m e^{4}}{8 h^{2} E_{0}^{2}} \frac{1}{m^{2}} - \frac{1}{n^{2}} Here, n \\ or \frac{h c}{λ} & = \frac{Z^{2} m e^{4}}{8 h^{2} E_{0}^{2}} \frac{1}{m^{2}} - \frac{1}{n^{2}} \Rightarrow \frac{1}{λ} \propto \frac{1}{m^{2}} - \frac{1}{n^{2}} Z^{2} \end{aligned}$

For first case,

$\begin{aligned} λ = 660 n m, m = 2 and n = 3 \\ ∴ \frac{1}{660 n m} \propto \frac{1}{(2)^{2}} - \frac{1}{(3)^{2}} Z^{2} \\ \Rightarrow \frac{1}{660 n m} \propto \frac{1}{4} - \frac{1}{9} Z^{2} or \frac{5}{36} Z^{2} \end{aligned}$

For second case, transition is from $n = 4$ to $n = 2$ , i.e. $m = 2$ and $n = 4$

$\begin{array}{ll} ∴ & \frac{1}{λ} \propto \frac{1}{(2)^{2}} - \frac{1}{(4)^{2}} Z^{2} \Rightarrow \frac{1}{λ} \propto \frac{1}{4} - \frac{1}{16} Z^{2} \\ or & \frac{1}{λ} \propto \frac{3}{16} Z^{2} \end{array}$

From Eqs. (i) and (ii), we get

$\begin{aligned} \frac{λ}{660 n m} & = \frac{5}{36} \times \frac{16}{3} \\ \Rightarrow λ & = \frac{80}{108} \times 660 n m = 488.9 n m \end{aligned}$

Modern Physics 1 Question 6

5. Taking the wavelength of first Balmer line in hydrogen spectrum $(n = 3$ to $n = 2)$ as $660 n m$ , the wavelength of the $2^{nd}$ Balmer line $(n = 4$ to $n = 2)$ will be

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Modern Physics 1 Question 6

5. Taking the wavelength of first Balmer line in hydrogen spectrum (n=3 to n=2) as 660nm, the wavelength of the 2nd Balmer line (n=4 to n=2) will be

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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5. Taking the wavelength of first Balmer line in hydrogen spectrum $(n = 3$ to $n = 2)$ as $660 n m$ , the wavelength of the $2^{nd}$ Balmer line $(n = 4$ to $n = 2)$ will be