Hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength 970.6 mathringA. How many different lines are possible in the resulting emission spectrum?

Question

Hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength 970.6 mathringA. How many different lines are possible in the resulting emission spectrum? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Six different lines (1/970.6 × 10-8)=109678 × 1((1/1)-(1/ n 22)) (9.1176 × 10-6/970.6 × 10-8)=1-(1/ n 22) ; (1/ n 22) =0.0606 n2=4 Number of lines =((4-1)(4)/2)=6

Q. Hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength $970.6 \overset{˚}{A}$ . How many different lines are possible in the resulting emission spectrum?

Six different lines
$\frac{1}{970.6 \times 1 0 ^{- 8}} = 109678 \times 1 (\frac{1}{1} - \frac{1}{n _{2}^{2}})$
$\frac{9.1176 \times 1 0 ^{- 6}}{970.6 \times 1 0 ^{- 8}} = 1 - \frac{1}{n _{2}^{2}}; \frac{1}{n _{2}^{2}}$
$= 0.0606$
$n_{2} = 4$
Number of lines $= \frac{( 4 - 1 ) ( 4 )}{2} = 6$

Q. Hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength 970.6A˚. How many different lines are possible in the resulting emission spectrum?

Six different lines 970.6×10−81​=109678×1(11​−n22​1​) 970.6×10−89.1176×10−6​=1−n22​1​;n22​1​ =0.0606 n2​=4 Number of lines =2(4−1)(4)​=6

Q. Hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength $970.6 \overset{˚}{A}$ . How many different lines are possible in the resulting emission spectrum?

Six different lines
$\frac{1}{970.6 \times 1 0 ^{- 8}} = 109678 \times 1 (\frac{1}{1} - \frac{1}{n _{2}^{2}})$
$\frac{9.1176 \times 1 0 ^{- 6}}{970.6 \times 1 0 ^{- 8}} = 1 - \frac{1}{n _{2}^{2}}; \frac{1}{n _{2}^{2}}$
$= 0.0606$
$n_{2} = 4$
Number of lines $= \frac{( 4 - 1 ) ( 4 )}{2} = 6$