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Question Number 52991 by prakash jain last updated on 16/Jan/19

Commented by prakash jain last updated on 16/Jan/19

Correct option is 2. NTA official key has 3. Please feedback. This question is also from 9th Jan JEE main

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$itismyopinion...$$$$\frac{1}{\lambda }=R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2})$$$$\stackrel{-}{\nu }=R_H(\frac{1}{8^2}-\frac{1}{n^2})$$$$alongxaxis\to \frac{1}{n^2}$$$$alongyaxis\to \stackrel{-}{\nu }$$$$soy=R_H(\frac{1}{8^2}-x)$$$$y=(-R_H)x+\frac{R_H}{8^2}$$$$y=mx+c$$$$soslopem=(-R_H)anditisstraightline$$$$sooption3iscorrect...$$

Commented by prakash jain last updated on 16/Jan/19

$$\frac{1}{\lambda }=\mathrm{R}[\frac{1}{n_f^2}-\frac{1}{n_i^2}]$$$$\mathrm{I}\mathrm{did}\mathrm{a}\mathrm{google}\mathrm{search}.$$

Commented by prakash jain last updated on 16/Jan/19

$$\mathrm{Maybe}\mathrm{the}\mathrm{question}\mathrm{could}\mathrm{be}\mathrm{wrong}$$$$\mathrm{imterpreation}\mathrm{issue}.$$

Commented by prakash jain last updated on 16/Jan/19

$$\mathrm{Can}\mathrm{you}\mathrm{check}\mathrm{the}\mathrm{last}\mathrm{question}$$$$\mathrm{as}\mathrm{well}.$$

Commented by prakash jain last updated on 16/Jan/19

Also the question says 8 to n. for example Balmer series emission spectrum is (1/2^2 - 1/n^2). From n to 2. So from 8 to n will mean n has to smaller than 8 for it to be emission spectrum.

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$plsuploadquestionpls...whichquestion...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

Commented by prakash jain last updated on 16/Jan/19

$$\mathrm{Question}52993$$

Commented by prakash jain last updated on 16/Jan/19

$$\mathrm{As}\mathrm{posted}\mathrm{in}\mathrm{your}\mathrm{material}\mathrm{balmer}$$$$\mathrm{series}\mathrm{is}n\mathrm{to}2\left(seearrows\right).\mathrm{So}\mathrm{the}\mathrm{formula}\mathrm{is}$$$$\frac{1}{2^2}-\frac{1}{{\mathrm{n}}^2}.$$$$\mathrm{For}8\mathrm{to}n\mathrm{the}\mathrm{formula}\mathrm{should}\mathrm{be}$$$$\frac{1}{n^2}-\frac{1}{8^2}.$$$$\mathrm{Am}\mathrm{I}\mathrm{wrong}?$$

Commented by prakash jain last updated on 16/Jan/19

$$n\mathrm{has}\mathrm{to}\mathrm{be}\mathrm{lower}\mathrm{than}8\mathrm{for}\mathrm{it}\mathrm{to}\mathrm{be}$$$$\mathrm{emission}\mathrm{spectrum}\mathrm{and}\mathrm{wavelength}$$$$\mathrm{cannot}\mathrm{be}\mathrm{negative}.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$oksirletmeconsultotherbook...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$Q52993belongstochemistry...butiamweakinit..$$

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