Grade 12Modern Physics

The two longest wavelengths of Balmer series of triply ionized beryllium (z=4) are:A)41nm B)30.4nm C)45nm D)39nmIt is a multi correct answer question. Please answer this it is very URGENT

Naga sree harsha reddy

8 Years agoGrade 12

2 Answers

Arun

8 Years ago

Dear student

Since the radiation of the Balmer series ends with n= 2.

Hence two longest wavelengths are the radiations corresponding to

n = 3 to n= 2

And n = 4 to n =2

Now find (E3- E2) = - (13.6 ev) 4² ( 1/9 - 1/4)

Then

Lambda = h c /E

Similarily for

(E4- E2)

Hope you understand.

Regards

Arun (askIITians forum expert)

kipkoech patrick

7 Years ago

1 first wavelength

n = 3 , E = _ 13.6 eV4 ²(1/9 – 1/4)

but lamda = hc/E

This gives the first longest wavelength .

for the second longest wavelenthg

take n = 4

therefore E = -13.6 eV 4² (1/16 – ¼)

Then lamda = hc/E

This gives the second wavelenthg

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The two longest wavelengths of Balmer series of triply ionized beryllium (z=4) are:A)41nm B)30.4nm C)45nm D)39nmIt is a multi correct answer question. Please answer this it is very URGENT

Naga sree harsha reddy

2 Answers

Arun

kipkoech patrick

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