The energy of electron in the nth orbit of hydrogen atom is expressed as En=(-13.6/n2) e V. The shortest and longest wavelength of Lyman series will be

Question

The energy of electron in the nth orbit of hydrogen atom is expressed as En=(-13.6/n2) e V. The shortest and longest wavelength of Lyman series will be (A) 910 mathringA , 1213 mathringA  (B) 5463 mathringA , 7858 mathringA  (C) 1315 mathringA , 1530 mathringA  (D) None of these

Tardigrade Admin · Accepted Answer

(1/λ max )=R[(1/(1)2)-(1/(2)2)] ⇒ λ max =(4/3 R) ≈ 1213 mathringA and (1/λ min )= R [(1/(1)2)-(1/∞)] ⇒ λ min =(1/ R ) ≈ 910 mathringA

Q. The energy of electron in the nth orbit of hydrogen atom is expressed as $E_{n} = \frac{- 13.6}{n ^{2}} e V$ . The shortest and longest wavelength of Lyman series will be

$910 \overset{˚}{A}, 1213 \overset{˚}{A}$

$5463 \overset{˚}{A}, 7858 \overset{˚}{A}$

$1315 \overset{˚}{A}, 1530 \overset{˚}{A}$

None of these

$\frac{1}{λ _{m a x}} = R [\frac{1}{( 1 ) ^{2}} - \frac{1}{( 2 ) ^{2}}] \Rightarrow λ_{m a x} = \frac{4}{3 R} \approx 1213 \overset{˚}{A}$
and $\frac{1}{λ _{m i n}} = R [\frac{1}{( 1 ) ^{2}} - \frac{1}{\infty}] \Rightarrow λ_{m i n} = \frac{1}{R} \approx 910 \overset{˚}{A}$

Q. The energy of electron in the nth orbit of hydrogen atom is expressed as En​=n2−13.6​eV. The shortest and longest wavelength of Lyman series will be

910A˚,1213A˚

5463A˚,7858A˚

1315A˚,1530A˚