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#ground state

10 public questions tagged with this topic.

De Broglie Hypothesis and Quantization in Bohr Model

What is the minimum energy required to excite a hydrogen atom from its ground state to the first excited state? (Use \(

**De Broglie hypothesis** λ = h/p, p=mv momentum, suggests electron as wave, in Bohr model circumference 2πr = n λ, standing wave condition, n wavelengths fit into orbit, for n=6, 6 wavelengths, for n=4, 4 wavelengths, explains quantization of angular momentum L = r p = r h/λ = r h n/(2πr)= n h/2π = n ħ, physical basis for Bohr quantization. E₁ = -13.6 eV , E₂ = -3.4 eV . Δ E = -3.4 - (-13.6) = 10.2 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

Ref: NCERT > Physics Book > Atoms and Nuclei > De Broglie Hypothesis and Quantization in Bohr Model

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Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of the electron in the ground state is -13.6 eV. What is its kinetic energy?

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. Total energy E = K + U , where K = (e²/8πepsilon₀ r) , U = -(e²/4πepsilon₀ r) . E = K - 2K = -K . E = -13.6 eV = -K ⇒ K = 13.6 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R =

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of an electron in the ground state is -13.6 eV. What is the magnitude of its potent

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. E = K + U , K = 13.6 eV , U = -2K = -27.2 eV . Magnitude = |U| = 27.2 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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Bohr Model Energy Levels and Hydrogen Spectrum

A hydrogen atom in the ground state is bombarded with a 12.5 eV electron beam. What is the maximum energy level the elec

**Excitation** energy required to go from n=1 to n=3 is 12.09 eV, from ground to n=∞ ionization 13.6 eV, state n=∞ means ionized, electron free with zero energy, highest level reached by electron beam energy determines which levels can be excited, e.g., 11 eV beam from ground can reach n=2 (10.2 eV) but not n=3 (12.09 eV), so max n=2. E₁ = -13.6 eV . E_n = -13.6 + 12.5 = -1.1 eV . -1.1 = -(13.6/n²) ⇒ n² ≈ 12.36 ⇒ n = 3 (since E₃ = -1.51 eV < -1.1 eV , n = 3 ). Using E_n = -13.6/n² eV, r_n =

Ref: NCERT > Physics Book > Atoms and Nuclei > Bohr Model Energy Levels and Hydrogen Spectrum

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Atomic Models - Rutherford, Thomson and Bohr

A hydrogen atom in the \( n = 5 \) state emits a photon and returns to the ground state. What is the maximum energy of t

**Thomson's plum pudding model** positive charge uniformly distributed in sphere with electrons embedded, positive charge spread, fails to explain large angle scattering observed. Bohr's model introduces stationary orbits with quantized angular momentum L = n h/2π, physical basis de Broglie standing wave condition 2πr = n λ, circumference fits n wavelengths, explains line spectrum. E₅ = -0.544 eV , E₁ = -13.6 eV . Δ E = -0.544 - (-13.6) = 13.056 eV ≈ 13.06 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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Hydrogen Atom Properties - Radius, Speed and Energy

What is the orbital period of an electron in the first orbit of a hydrogen atom if its speed is \( 2.2 \times 10^6 \, \t

**Angular momentum** L_n = n h/2π, h=6.6×10⁻³⁴ J·s, L₂=2×6.6×10⁻³⁴/2π=2.11×10⁻³⁴ J·s, speed v_n = e²/(2 ε₀ h) ×1/n ≈2.2×10⁶/n m/s, kinetic energy ½ m v² =13.6/n² eV, illustrating Bohr model predictions for hydrogen-like atoms. T = (2π r/v) = (2 × 3.14 × 5.3 × 10⁻¹¹/2.2 × 10⁶) ≈ 1.51 × 10⁻¹⁶ s . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 1.51 × 10⁻¹⁶ s, consistent with Bohr model and nuclear binding energy systematics.

Ref: NCERT > Physics Book > Atoms and Nuclei > Hydrogen Atom Properties - Radius, Speed and Energy

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