Question

Wavelength for high energy EMR transition in H-atom is 91 nm. What energy is needed for this
transition?
(a) 1.36 eV
(b) 1240 eV
(c) 13 eV
(d) 13.6 eV​

Answer 1

Hello Dear,

◆ Answer -

(d) 13.6 eV

◆ Explaination -

Energy required for transition of electron in H-atom is given by -

E = hc/λ

E = 6.63×10^-34 × 3×10^8 / 91×10^-9

E = 2.186×10^-18 J

Our options are in eV. So let's convert this to eV.

E = 2.186×10^-18 / 1.6×10^-19

E = 13.66 eV

Therefore, energy needed for transition is 13.66 eV.

Thanks dear...

Answer 2

$\huge \underline \bold \color{coral} \mathfrak{Answer }$

$\lambda = 91nm = 91 \times {10}^{ - 9} m$

$E = \frac{hc}{ \lambda}$

Where h is Planck's constant, c is speed of light in vaccum and lambda is wavelength.

$E = \frac{6.63 \times 10 {}^{ - 34 } \times 3 \times {10}^{8} }{91 \times {10}^{ - 9} }$

$E = 2.186 \times 10 {}^{ - 18} J$

$1 \: Joule = 1.6 \times {10}^{ - 19} ev$

$E = 2.186 \times {10}^{ - 18} \times 1.6 \times {10}^{ - 19} ev$

$\huge{E} = \huge \red {13.66 \: ev}$

Hope it helps....❤️