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(a) Using data from the Table, find the mass destroyed when the energy in a barrel of crude oil is released. (b) Given these barrels contain 200 liters and assuming the density of crude oil is 750kg/m3,750 \mathrm{kg} / \mathrm{m}^{3}, what is the ratio of mass destroyed to original mass, Δm/m?\Delta m / m?

 Object/phenomenon  Energy in joules  Big Bang 1068 Energy released in a supernova 1044 Fusion of all the hydrogen in Earth’s oceans 1034 Annual world energy use 4×1020 Large fusion bomb (9 megaton) 3.8×1016 1 kg hydrogen (fusion to helium) 6.4×1014 1 kg uranium (nuclear fission) 8.0×1013 Hiroshima-size fission bomb (10 kiloton) 4.2×101390,000 -ton aircraft carrier at 30 knots 1.1×1010 1 barrel crude oil 5.9×1091 ton TNT4.2×1091 gallon of gasoline 1.2×108 Daily home electricity use (developed countries) 7×107 Daily adult food intake (recommended) 1.2×1071000kg car at 90km/h3.1×1051g fat (9.3kcal)3.9×104 ATP hydrolysis reaction 3.2×1041g carbohydrate (4.1kcal)1.7×1041g protein (4.1kcal)1.7×104 Tennis ball at 100km/h22 Mosquito (102 g at0.5m/s)1.3×106 single electron in a TV tube beam 4.0×1015 Energy to break one DNA strand 1019\begin{array}{|l|l|} \hline {\text { Object/phenomenon }} & \text { Energy in joules } \\ \hline \text { Big Bang } & 10^{68} \\ \hline \text { Energy released in a supernova } & 10^{44} \\ \hline \text { Fusion of all the hydrogen in Earth's oceans } & 10^{34} \\ \hline \text { Annual world energy use } & 4 \times 10^{20} \\ \hline \text { Large fusion bomb (9 megaton) } & 3.8 \times 10^{16} \\ \hline \text { 1 kg hydrogen (fusion to helium) } & 6.4 \times 10^{14} \\ \hline \text { 1 kg uranium (nuclear fission) } & 8.0 \times 10^{13} \\ \hline \text { Hiroshima-size fission bomb (10 kiloton) } & 4.2 \times 10^{13} \\ \hline 90,000 \text { -ton aircraft carrier at 30 knots } & 1.1 \times 10^{10} \\ \hline \text { 1 barrel crude oil } & 5.9 \times 10^{9} \\ \hline 1 \text { ton } \mathrm{TNT} & 4.2 \times 10^{9} \\ \hline 1 \text { gallon of gasoline } & 1.2 \times 10^{8} \\ \hline \text { Daily home electricity use (developed countries) } & 7 \times 10^{7} \\ \hline \text { Daily adult food intake (recommended) } & 1.2 \times 10^{7} \\ \hline 1000-\mathrm{kg} \text { car at } 90 \mathrm{km} / \mathrm{h} & 3.1 \times 10^{5} \\ \hline 1 \mathrm{g} \text { fat }(9.3 \mathrm{kcal}) & 3.9 \times 10^{4} \\ \hline \text { ATP hydrolysis reaction } & 3.2 \times 10^{4} \\ \hline 1 \mathrm{g} \text { carbohydrate }(4.1 \mathrm{kcal}) & 1.7 \times 10^{4} \\ \hline 1 \mathrm{g} \text { protein }(4.1 \mathrm{kcal}) & 1.7 \times 10^{4} \\ \hline \text { Tennis ball at } 100 \mathrm{km} / \mathrm{h} & 22 \\ \hline \text { Mosquito }\left(10^{-2}\ \mathrm{g}\ \mathrm{at} 0.5 \mathrm{m} / \mathrm{s}\right) & 1.3 \times 10^{-6} \\ \hline \text { single electron in a TV tube beam } & 4.0 \times 10^{-15} \\ \hline \text { Energy to break one DNA strand } & 10^{-19} \\ \hline \end{array}

Question

A singly ionized helium ion has only one electron and is denoted He+.\mathrm{He}^{+}. What is the ion’s radius in the ground state compared to the Bohr radius of hydrogen atom?

Solution

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Concept: \textbf{Concept: }

As we know atom is the smallest unit of elements which is combine to form molecules and the molecules are combining to form the matter. Also, we know that the first direct observation of atoms was in brownian motion which give accurate sizes for atoms in range 1010 m10^{-10} \mathrm{~m}. Also, we know that the atom consists of electrons, protons and neutrons. We know that the charge-to-mass ratios are given by

qeme=1.76×1011 C/kgqpmp=9.57×107 C/kg\begin{align*} \dfrac{q_{e} }{ m_{e} } &= - 1.76 \times 10^{11} \mathrm{~C/kg} \\ \dfrac{ q_{p} }{ m_{p} } &= 9.57 \times 10^{7} \mathrm{~C/kg} \\ \end{align*}

Where, the mass of the electron is me=9.11×1031 kgm_{e} = 9.11 \times 10^{-31} \mathrm{~kg} and the mass of proton is mp=1.67×1027 kgm_{p} = 1.67 \times 10^{-27} \mathrm{~kg}.

According to Bohr's theory of Hydrogen atom, atomic and molecular spectra are quantized with hydrogen spectrum wavelength which is given by

1λ=R (1nf21ni2)\begin{align*} \dfrac{1}{ \lambda } &= R ~ \left( \dfrac{1}{ n_{f}^{2} } - \dfrac{1}{ n_{i}^{2} } \right) \end{align*}

Where, λ\lambda is the wavelength of the emitted electromagnetic radiation and R=1.097×107 m1R = 1.097 \times 10^{7} \mathrm{~m^{-1}} is the Rydberg constant.

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