An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg 's constant R = 105 cm-1 the frequency in hertz of the emitted radiation will be

Question

An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg 's constant R = 105 cm-1 the frequency in hertz of the emitted radiation will be (A)  (3/16)× 105  (B)  (3/16)× 1015  (C)  (9/16)× 1015  (D)  (3/4)× 1015

Tardigrade Admin · Accepted Answer

Frequency, v =(c/λ)=c ⋅ R( (1/n12)-(1/n22) ) =3 × 108 × 107( (1/22)-(1/42)) =(9/16) × 1015 Hz

Q. An electron jumps from the $4^{t h}$ orbit to $2^{n d}$ orbit of hydrogen atom. Given the Rydberg 's constant $R = 1 0^{5} c m^{- 1}$ the frequency in hertz of the emitted radiation will be

$\frac{3}{16} \times 1 0^{5}$

$\frac{3}{16} \times 1 0^{15}$

$\frac{9}{16} \times 1 0^{15}$

$\frac{3}{4} \times 1 0^{15}$

Frequency, $v = \frac{c}{λ} = c \cdot R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$= 3 \times 1 0^{8} \times 1 0^{7} (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}})$
$= \frac{9}{16} \times 1 0^{15} Hz$

Q. An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg 's constant R=105cm−1 the frequency in hertz of the emitted radiation will be

163​×105

163​×1015

169​×1015

43​×1015