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Question: Based on the equation: \(\Delta E=-2.0\times {{10}^{-18}}J\) \((\dfrac{1}{{{n}^{2}}_{2}}-\dfrac{1}{{...

Based on the equation: ΔE=−2.0×10−18J\Delta E=-2.0\times {{10}^{-18}}J (1n22−1n21)(\dfrac{1}{{{n}^{2}}_{2}}-\dfrac{1}{{{n}^{2}}_{1}}), the wavelength of the light that must be absorbed to excite hydrogen electron from level n=1 to level n=2 will be: (h = 6.625×10−346.625\times {{10}^{-34}}J s, C = 3×108ms−13\times {{10}^{8}}m{{s}^{-1}})
A. 2.650×10−7m2.650\times {{10}^{-7}}m
B. 1.325×10−7m1.325\times {{10}^{-7}}m
C. 5.300×10−10m5.300\times {{10}^{-10}}m
D. 1.325×10−10m1.325\times {{10}^{-10}}m

Explanation

Solution

Hydrogen is the first element of the periodic table represented by the symbol H and has atomic number 1 and atomic weight of 1.008, hydrogen is the lightest element in the periodic table. In the universe hydrogen is the most abundant chemical substance.

Complete Step by step solution: At STP i.e. standard temperature and pressure, hydrogen is present in the form of diatomic colorless, odorless, tasteless, non-toxic, nonmetallic highly combustible gas with shows the molecular formula H2{{H}_{2}}. With most nonmetallic elements hydrogen forms covalent compounds. There is an important role of hydrogen in an acid–base reactions because mostly acid-base reactions involve the exchange of protons between soluble molecules. In ionic compounds hydrogen is found with negative charge where it is known by a hydride or can be found as a positively charged species denoted by the symbol H+{{H}^{+}}.
ΔE=−2.0×10−18J\Delta E=-2.0\times {{10}^{-18}}J; Wavelength can be calculated by applying formula
1λ=RZ2(1n12−1n22)\dfrac{1}{\lambda }=R{{Z}^{2}}(\dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{2}}^{2}})
Where R corresponds to Rydberg constant having value 1.025×10−71.025\times {{10}^{-7}}m and Z = 1 in case of hydrogen and n1=1{{n}_{1}}=1and n2=2{{n}_{2}}=2; substitute all the values in equation
1λ=1.025×10−7(112−122)\dfrac{1}{\lambda }=1.025\times {{10}^{-7}}(\dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}}); 1λ=1.025×10−7(34)=0.768×10−7\dfrac{1}{\lambda }=1.025\times {{10}^{-7}}(\dfrac{3}{4})=0.768\times {{10}^{-7}}
λ=1.30×10−7\lambda =1.30\times {{10}^{-7}}

Hence, option B is correct.

Note: The Rydberg formula is a mathematical formula which is used to find the wavelength of light resulting from an electron moving between energy levels of an atom. The energy of an electron changes when it goes to one atomic orbital to another atomic orbital. A photon of light is created when the electron changes from an orbital with high energy to a lower energy state and when it moves from low energy to a higher energy state a photon of light is absorbed by the atom.