Question

energy of the electron in orbit of h atom is - 1.51 electron volt the wavelength produced by the electron in the same Orbit if first orbit of h have radius x​

Answer 1

Answer:

The wavelength produced by the electron in the given orbit is 1,235.9 nm.

Explanation:

Energy of the electron in the given orbit of hydrogen atom = E =-1.51 eV

$E = -1.51 eV=-1.51\times 1.60218\times 10^{-19}J=-1.6039\times 10^{-19} J$

$1 eV=1.60218\times 10^{-19} Joules$

$E=\frac{h\times c}{\lambda}$

where,

E = energy of photon of an emitted wavelength

h = Planck's constant = $6.63\times 10^{-34}Js$

c = speed of light = $6.63\times 10^{-34}Js$

$\lambda$ = wavelength of light emitted

$E=1.6039\times 10^{-19} J$

$\lambda = \frac{6.63\times 10^{-34}Js\times 6.63\times 10^{-34}Js}{1.6039\times 10^{-19} J}$

$\lambda =1.2359\times 10^{-6} m=1,235.9 nm$

The wavelength produced by the electron in the given orbit is 1,235.9 nm.

Answer 2

Answer:

The wavelength produced by the electron in the same orbit if first orbit of h have radius x​ is 821 nm.

Explanation:

$1eV=1.602 \times 10^{-19} J$

So, Energy

$E = 1.51eV \times \frac{(1.602 \times 10^{-19} J)}{1eV}=2.42 \times 10^{-19} J$

$E=h \times v\\\\E=h \times \frac {c}{\lambda}\\\\\lambda =h \times \frac {c}{E}$

$\lambda = \frac {(6.626 \times 10^{-34} J.s \times 3 \times 10^8 m s^{-1}))}{(2.42 \times 10^{-19} J)}$

$\lambda =8.21 \times 10^{-7} m$

$1m=10^9 nm$

So,

$\lambda = 8.21 \times 10^{-7} m \times \frac{10^9 nm}{1m}$

$\lambda =821 nm$

(Answer)

Please note :

E stands for energy  

c for speed of light

$\lambda$ for wavelength

$f$ for frequency