Q.

For the Balmer series in the spectrum of H atom,v-=RH1n12-1n22the correct statements
among (I) to (IV) are:

I) As wavelength decreases, the lines in the
series converge
II) The integer n1 is equal to 2

III) The lines of longest wavelength corresponds to n2 = 3
IV) The ionization energy of hydrogen can be calculated from wave number of these lines

 

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

a

(II), (III), (IV)

b

(I), (II), (III)

c

(I), (II), (IV)

d

(I), (III), (IV)

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

v-=RH1n12-1n22

Here n=2 since Balmer series, the transition is between some energy level  n2  

and n1=2.

As the wave length decreases, the wave number increases. This implies that the difference122-1n22also increases or ultimately n2 increases. As n2increases , the lines in the series converge. Ionization energy of hydrogen can be calculated from wave number of lines in lyman series as the electron is present in n=1.Longest wave length.

corresponds to minimum difference betweeen 122 and 1n22 this is for n2=3 

 

Watch 3-min video & get full concept clarity
AITS_Test_Package
AITS_Test_Package
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon