the limiting line of balmer series lies on?
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the limiting line of balmer
series is in visible region
series is in visible region
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1
ANSWER:
METHOD 1:
For limiting line of Balmer series, n1=2 and n2 =3
v =RH/ h (1/n12 - 1/n22)
= 3.29×1015(1/4 - 1/ 9) Hz
= 4.57 × 1014 Hz
METHOD 2
Explanation:
The Balmer series corresponds to all electron transitions from a higher energy level to n=2.
The wavelength is given by the Rydberg formula
where
R= the Rydberg constant and
n1 and n2 are the energy levels such that n2>n1
Since fλ=c
we can re-write the equation as
1λ=fc=R(1n21−1n22)
or
f=cR(1n21−1n22)=R′(1n21−1n22)
where R′ is the Rydberg constant expressed in energy units (3.290×1015lHz).
In this problem, n1=2, and the frequency of the limiting line is reached
as n→∞.
Thus,
f=limn→∞R′(14−1n22)=R′(0.25−0)=0.25R′
=0.25×3.290×1015lHz=8.225×1014lHz
METHOD 1:
For limiting line of Balmer series, n1=2 and n2 =3
v =RH/ h (1/n12 - 1/n22)
= 3.29×1015(1/4 - 1/ 9) Hz
= 4.57 × 1014 Hz
METHOD 2
Explanation:
The Balmer series corresponds to all electron transitions from a higher energy level to n=2.
The wavelength is given by the Rydberg formula
where
R= the Rydberg constant and
n1 and n2 are the energy levels such that n2>n1
Since fλ=c
we can re-write the equation as
1λ=fc=R(1n21−1n22)
or
f=cR(1n21−1n22)=R′(1n21−1n22)
where R′ is the Rydberg constant expressed in energy units (3.290×1015lHz).
In this problem, n1=2, and the frequency of the limiting line is reached
as n→∞.
Thus,
f=limn→∞R′(14−1n22)=R′(0.25−0)=0.25R′
=0.25×3.290×1015lHz=8.225×1014lHz
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