The shortest wavelength of H atom in the Lyman series is λ₁. The longest wavelength in the Balmer Series of H⁺ is:

Option 1 - <p>5λ₁/9<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>36λ₁/5<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>9λ₁/5<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>27λ₁/5<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
3 Views|Posted 5 months ago
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1 Answer
V
5 months ago
Correct Option - 3
Detailed Solution:

For hydrogen atom :
For Lyman series n? = 1 & n? = ∞
1/λH = RH [1/1² - 1/∞²] So, λ = 1/RH
For He? ion Balmer series n? = 2 & n? = 3
1/λHe? = RH * Z² [1/n? ² - 1/n? ²]
1/λHe? = RH * 4 * [1/4 - 1/9] = RH * 4 * (5/36)
1/λHe? = (5/9)RH = (5/9) (1/λ)
(λHe? ) = (9/5)λ

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