If the energy difference between the ground state and excited state of an atom is 4.4 × 10-19 J . The wavelength of photon required to produce this transition is

Question

If the energy difference between the ground state and excited state of an atom is 4.4 × 10-19 J . The wavelength of photon required to produce this transition is (A) 4.5 × 10-7 m (B) 4.5 × 10-7 nm (C) 4.5 × 10-7 A (D) 4.5 × 10-7 cm

Tardigrade Admin · Accepted Answer

Δ E =[Excited state - ground state ]=4.4 × 10-19 J According to Plank's quantum theory longrightarrow Δ E =( nhc /λ) λ=( hc /Δ E ) =(6.6 × 10-34 J - s × 3 × 108 m / s /4.4 × 10-19 J ) =4.5 × 10-7 m

Q. If the energy difference between the ground state and excited state of an atom is $4.4 \times 1 0^{- 19} J .$ The wavelength of photon required to produce this transition is

$4.5 \times 1 0^{- 7} m$

$4.5 \times 1 0^{- 7} nm$

$4.5 \times 1 0^{- 7} A$

$4.5 \times 1 0^{- 7} c m$

$Δ E =$ [Excited state $-$ ground state $] = 4.4 \times 1 0^{- 19} J$
According to Plank's quantum theory $⟶ Δ E = \frac{nh c}{λ}$
$λ = \frac{h c}{Δ E}$
$= \frac{6.6 \times 1 0 ^{- 34} J - s \times 3 \times 1 0 ^{8} m / s}{4.4 \times 1 0 ^{- 19} J}$
$= 4.5 \times 1 0^{- 7} m$

Q. If the energy difference between the ground state and excited state of an atom is 4.4×10−19J. The wavelength of photon required to produce this transition is

4.5×10−7m

4.5×10−7nm

4.5×10−7A

4.5×10−7cm