The frequency of light emitted, when the electron makes transition from the level of principle quantum number n=2 to the level with n=1 is (Take, the ionization energy of hydrogen to be 13.6 eV and .h ≃ 4 × 10-15 eV - s )

Question

The frequency of light emitted, when the electron makes transition from the level of principle quantum number n=2 to the level with n=1 is (Take, the ionization energy of hydrogen to be 13.6 eV and .h ≃ 4 × 10-15 eV - s ) (A) 2.55 × 1015 Hz (B) 1.7 × 1015 Hz (C) 3.4 × 1015 Hz (D) 5.1 × 1015 Hz

Tardigrade Admin · Accepted Answer

As ionisation energy is 13.6 eV, so its negative will be the energy of atom in ground state (n=1) hence E1=-13.6 eV Energy in second orbit (n=1) E2=-(13.6/22)=-34 eV When electron jumps from n=2 to n=1, then energy of photon emitted by atom Δ E =E2-E1 =-3.4-(-13.6) Δ E =10.2 eV So, frequency (v), h v =10.2 eV v =(10.2 eV /h)=(10.2 eV /4 × 10-15 eV - s ) v =255 × 10+15 Hz

Q. The frequency of light emitted, when the electron makes transition from the level of principle quantum number $n = 2$ to the level with $n = 1$ is (Take, the ionization energy of hydrogen to be $13.6 e V$ and $h ≃ 4 \times 1 0^{- 15} e V - s)$

$2.55 \times 1 0^{15} Hz$

$1.7 \times 1 0^{15} Hz$

$3.4 \times 1 0^{15} Hz$

$5.1 \times 1 0^{15} Hz$

Q. The frequency of light emitted, when the electron makes transition from the level of principle quantum number n=2 to the level with n=1 is (Take, the ionization energy of hydrogen to be 13.6eV and h≃4×10−15eV−s)

2.55×1015Hz

1.7×1015Hz

3.4×1015Hz

5.1×1015Hz

Q. The frequency of light emitted, when the electron makes transition from the level of principle quantum number $n = 2$ to the level with $n = 1$ is (Take, the ionization energy of hydrogen to be $13.6 e V$ and $h ≃ 4 \times 1 0^{- 15} e V - s)$

$2.55 \times 1 0^{15} Hz$

$1.7 \times 1 0^{15} Hz$

$3.4 \times 1 0^{15} Hz$

$5.1 \times 1 0^{15} Hz$