Question

the frequency of second line of Balmer series in spectrum of Be^ +3​

Answer 1

Explanation:

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Answer 2

Answer:

the frequency in spectrum of $Be^{+3}$ for the Balmer series is $9.9*10^{15}$

Explanation:

Balmer series is obtained when transition of electron takes place from higher principal quantum numbers to the 2nd quantum number.

From n>2 to n=2

The wavenumber (1/λ) is given the Rydberg equation,

1/λ $=Rz^{2} (\frac{1}{n_{1} ^{2} }- \frac{1}{n_{2} ^{2} } )$

where λ= wavelength

R= Rydberg constant≈1.1×10⁷$m^{-1}$

n₁=2(for Balmer series)

n₂=(n)principal quantum number from where transition takes place

For the second line of Balmer series of $Be^{+3}$ ion is

Z=4

n₂=4

1/ λ=$1.1*10^{7} *4^{2} (\frac{1}{2^{2} }-\frac{1}{4^{2} } )$

On solving,

1/λ=3.3*10⁷

Frequency and wavelength are related by a standard formula,

$Frequency=\frac{Speed\ of\ light}{wavelength\ of\ light}$

$f=3*10^{8} *3.3*10^{7} \\=9.9*10^{15} Hz$

Hence the frequency in spectrum of $Be^{+3}$ for the Balmer series is $9.9*10^{15}$