If the wavelength of the series limit of Lyman series for He⁺ ion is xA⁰,then what will be the wavelength of the series limit of Balmer series for Li⁺⁺ ion?
A) 9x/4 A⁰
B) 16x/9 A⁰
C) 5x/4 A⁰
D) 4x/7 A⁰
Show the appropriate calculation.Lazy answers will be deleted
Dhruv00:
Bella whats the answer?
my answer is in the order of B
Answers
Answered by
92
According to Rydberg formula,
1/lemda = RZ²[ 1/n1² - n2²]
where R is Rydberg constant .
for any series limit n2 = ∞
then , 1/lemda = RZ²[ 1/n1² - 1/∞]
1/lemda = RZ²/n1²
1/lemda = R( Z/n1)²
hence, wavelength is inversely proportional to square of Atomic no and directly proportional to square of n1 .
hence,
lemda1/lemda2 = ( Z2/Z1)²( n1/n2)²
here,
Z2 = 3
Z1 = 2
n1 = 1
n2 = 2
lemda1 = x A°
now,
Lemda1/lemda2 = (3/2)² × ( 1/2)² = 9/16
xA°/lemda2 = 9/16
lemda 2 = 16x/9 A°
so, option B is correct .
1/lemda = RZ²[ 1/n1² - n2²]
where R is Rydberg constant .
for any series limit n2 = ∞
then , 1/lemda = RZ²[ 1/n1² - 1/∞]
1/lemda = RZ²/n1²
1/lemda = R( Z/n1)²
hence, wavelength is inversely proportional to square of Atomic no and directly proportional to square of n1 .
hence,
lemda1/lemda2 = ( Z2/Z1)²( n1/n2)²
here,
Z2 = 3
Z1 = 2
n1 = 1
n2 = 2
lemda1 = x A°
now,
Lemda1/lemda2 = (3/2)² × ( 1/2)² = 9/16
xA°/lemda2 = 9/16
lemda 2 = 16x/9 A°
so, option B is correct .
bella i hope this will correct
you got it ,
??
Answered by
11
ANSWER IS NO.B)16×/9 A°............
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