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4 years ago

the ionization energies of several of the Group 1 elements in the periodic table are shown in the table below

1 answer:

3 0

In a group,the ionization energy decreases because the atomic size increases(no. of shells increases, therefore due to this, there's a weak attractive force between nucleus and the valence electrons)

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calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

Answer:

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.

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3 years ago

What kind of bond is N and C

yKpoI14uk [10]

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3 years ago

Need help with this

DochEvi [55]

A) 2, 3 , 2 simply valence the rest

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Write two ways to reduce carbon dioxide in atmosphere which result to global warming.

SCORPION-xisa [38]

Answer:

1- Using buses rather than opting for cars.

2- Regularly servicing your cars.

Explanation:

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3 years ago

H2CO3 -> H2O + CO2 What type of chemical reaction is taking place in this

OlgaM077 [116]

Answer: decomposition

Explanation: one molecule forms tro molecules

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3 years ago