Electrons having kinetic energy 30 eV are made to collide with atomic hydrogen gas (in ground state) and 42.5 % of electron energy is used to excite the hydrogen atoms. The number of emitted wavelengths in emission spectra are

Question

Electrons having kinetic energy 30 eV are made to collide with atomic hydrogen gas (in ground state) and 42.5 % of electron energy is used to excite the hydrogen atoms. The number of emitted wavelengths in emission spectra are (A) 3 (B) 6 (C) 12 (D) 18

Tardigrade Admin · Accepted Answer

42.5 % of 30 eV=12.75 eV ∴ E4-E1=-0.85-(- 13.6)=12.75 eV So, it means electron has been transmitted from ground state to 3r d excited state (n = 4) , so total number of unique wavelengths emitted are =(4 × (4 - 1)/2)=6

Q. Electrons having kinetic energy $30 e V$ are made to collide with atomic hydrogen gas (in ground state) and $42.5%$ of electron energy is used to excite the hydrogen atoms. The number of emitted wavelengths in emission spectra are

$3$

$6$

$12$

$18$

$42.5%$ of $30 e V = 12.75 e V$
$∴ E_{4} - E_{1} = - 0.85 - (- 13.6) = 12.75 e V$
So, it means electron has been transmitted from ground state to $3^{r d}$ excited state $(n = 4)$ , so total number of unique wavelengths emitted are
$= \frac{4 \times ( 4 - 1 )}{2} = 6$

Q. Electrons having kinetic energy 30eV are made to collide with atomic hydrogen gas (in ground state) and 42.5% of electron energy is used to excite the hydrogen atoms. The number of emitted wavelengths in emission spectra are

3

6

12

18

42.5% of 30eV=12.75eV ∴E4​−E1​=−0.85−(−13.6)=12.75eV So, it means electron has been transmitted from ground state to 3rd excited state (n=4) , so total number of unique wavelengths emitted are =24×(4−1)​=6

Q. Electrons having kinetic energy $30 e V$ are made to collide with atomic hydrogen gas (in ground state) and $42.5%$ of electron energy is used to excite the hydrogen atoms. The number of emitted wavelengths in emission spectra are

$3$

$6$

$12$

$18$

$42.5%$ of $30 e V = 12.75 e V$
$∴ E_{4} - E_{1} = - 0.85 - (- 13.6) = 12.75 e V$
So, it means electron has been transmitted from ground state to $3^{r d}$ excited state $(n = 4)$ , so total number of unique wavelengths emitted are
$= \frac{4 \times ( 4 - 1 )}{2} = 6$