Question

What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit? What is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 x 10^(-11) ergs.
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Answer 1

Explanation:

E₁ = - 2.18 × 10⁻¹¹ ergs

En = E₁ / n²

E₅ = (-2.18 × 10⁻¹¹) / 5²

E₅ = (-2.18 × 10⁻¹¹) / 25 = - 8.72 × 10⁻¹³

ΔE = E₅ - E₁

= - 8.72 × 10⁻¹³ - (- 2.18 × 10⁻¹¹) = 2.09828 × 10⁻¹¹ ergs.

1 ergs = 10⁻⁷ joules

Converting in joules :

2.09828 × 10⁻¹¹ × 10⁻⁷ = 2.09828 × 10⁻¹⁸ Joules

ΔE = hc/y

Y = wavelength

h = planks constant = 6.62 × 10⁻³⁴

c = speed of light = 3.0 × 10⁸

Doing the substitution we have :

2.09828 × 10⁻¹¹ = (6.62 × 10⁻³⁴ × 3 × 10⁸) / Y

Y = (6.62 × 10⁻³⁴ × 3 × 10⁸) / (2.09828 × 10⁻¹¹) = 9.465 × 10⁻¹⁵ cm

= 9.465 × 10⁻¹⁵ cm.

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Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Atomic - Radii related formulas have been used. We see that we are given the ground state electron energy of hydrogen. So firstly we can convert it into joules which is standard unit. Then we can find the required things by using formula.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{E_{n}\;=\;\bf{\dfrac{E_{1}}{n^{2}}}}}}$

$\\\;\boxed{\sf{\pink{\Delta\:E\;=\;\bf{E_{n_{1}}\;-\;E_{n_{2}}}}}}$

$\\\;\boxed{\sf{\pink{\Delta\:E\;=\;\bf{\dfrac{hc}{\lambda}}}}}$

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★ Solution :-

Given,

» Ground state electron energy = E₁ = -2.18 × 10⁻¹¹ ergs

» Planck's Constant = h = 6.6 × 10⁻³⁴ J sec

» Speed of light = c = 3 × 10⁸ m sec⁻¹

• Let the electron energy in fifth energy be E₅

• Let the nth orbit of hydrogen atom be denoted by n.

We know that,

→ 1 Joule = 10⁷ ergs

→ 1 erg = 10⁻⁷ Joules

→ -2.18 × 10⁻¹¹ ergs = -2.18 × 10⁻¹¹ × 10⁻⁷ Joules

→ -2.18 × 10⁻¹¹ ergs = -2.18 × 10⁻¹⁸ Joules

• For the value of E₅ ::

We know that,

$\\\;\sf{:\rightarrow\;\;E_{n}\;=\;\bf{\dfrac{E_{1}}{n^{2}}}}$

By replacing the value of n by 5 , we get

$\\\;\sf{:\rightarrow\;\;E_{5}\;=\;\bf{\dfrac{-2.18\:\times\:10^{-18}}{5^{2}}}}$

$\\\;\sf{:\rightarrow\;\;\green{E_{5}\;=\;\bf{8.72\:\times\:10^{-20}}}}$

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~ For the value of energy absorbed to shift electron from First to Fifth Bohr Orbit ::

This is given as,

$\\\;\sf{:\rightarrow\;\;\Delta\:E\;=\;\bf{E_{n_{1}}\;-\;E_{n_{2}}}}$

• Where, n₁ = First Orbit

• n₂ = Fifth Orbit

• ∆E denotes the energy absorbed.

Then,

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{\dfrac{-2.18\:\times\:10^{-18}}{5^{2}}\;-\;\dfrac{-2.18\:\times\:10^{-18}}{1^{2}}}}$

Taking -2.18 × 10⁻¹⁸ in common, we get

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{-2.18\:\times\:10^{-18}\bigg(\dfrac{1}{5^{2}}\;-\;\dfrac{1}{1^{2}}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{-2.18\:\times\:10^{-18}\bigg(\dfrac{1}{25}\;-\;\dfrac{1}{1}}\bigg)}$

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{-2.18\:\times\:10^{-18}\bigg(\dfrac{1\;-\;25}{25}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{-2.18\:\times\:10^{-18}\bigg(\dfrac{-\;24}{25}\bigg)}}$

Cancelling the -ve sign, we get

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{2.18\:\times\:10^{-18}\bigg(\dfrac{24}{25}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{2.18\:\times\:10^{-18}\:\times\:0.96}}$

$\\\;\sf{:\Longrightarrow\;\;\Delta\:E\;=\;\bf{\red{2.9\:\times\:10^{-18}\;\;Joules}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\:energy\;\:absorbed\;=\;\bf{\purple{2.9\:\times\:10^{-18}\;\;Joules}}}}}$

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~ For the wavelength of light emitted when electron returns to ground state ::

We know that ,

$\\\;\sf{:\rightarrow\;\;\Delta\:E\;=\;\bf{\dfrac{hc}{\lambda}}}$

• Here λ denotes the wavelength.

By applying values, we get

$\\\;\sf{:\rightarrow\;\;2.9\:\times\:10^{-18}\;=\;\bf{\dfrac{(6.6\:\times\:10^{-34})\:\times\:(3\:\times\:10^{8})}{\lambda}}}$

By replacing the values, we get

$\\\;\sf{:\rightarrow\;\;\lambda\;=\;\bf{\dfrac{(6.6\:\times\:10^{-34})\:\times\:(3\:\times\:10^{8})}{2.9\:\times\:10^{-18}}}}$

$\\\;\sf{:\rightarrow\;\;\lambda\;=\;\bf{956\;\times\;10^{-7}\;\;m}}$

$\\\;\sf{:\rightarrow\;\;\lambda\;=\;\bf{\orange{956\;\times\;10^{-7}\;\;m}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\:wavelength\;\:of\;\:light\;\:emitted\;=\;\bf{\purple{956\;\times\;10^{-7}\;\;m}}}}}$

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★ More formulas to know :-

$\\\;\sf{\leadsto\;\;\nu\;=\;\dfrac{\Delta\;E}{h}}$

$\\\;\sf{\leadsto\;\;E_{n}\;=\;-2.18\:\times\:10^{-18}\bigg(\dfrac{Z^{2}}{n^{2}}\bigg)J}$

$\\\;\sf{\leadsto\;\;\lambda\;=\;\dfrac{h}{mv}}$

$\;\sf{\leadsto\;\;\lambda\;=\;\dfrac{h}{p}}$

$\\\;\sf{\leadsto\;\;E_{n}\;=\;-\:R_{H}\bigg(\dfrac{1}{n^{2}}\bigg)}$