If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and (1/4pi 0) = 9 x 109 in SI units)

By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r0). Ki + Ui = Kf + Uf Initial kinetic energy Ki = 7.7 MeV = 7.7 x 106 x 1.6 x 10-19 J. Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z1e)(Z2e)/r0. Alpha particle has charge 2e. Class 12 Physics: Atoms

Solution & Explanation

### Related Formula

K_textinitial = U_textclosest approach

K = frac14pi epsilon_0 frac(2e)(Ze)r_0

### Core Logic By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (

r_0

K_i + U_i = K_f + U_f

K_i + 0 = 0 + frac14piepsilon_0 frac(2e)(79e)r_0

### Step 1: Convert Energy and Solve Initial kinetic energy

K_i = 7.7text MeV = 7.7 times 10^6 times 1.6 times 10^-19text J

7.7 times 10^6 times 1.6 times 10^-19 = frac9 times 10^9 times (2 times 1.6 times 10^-19) times (79 times 1.6 times 10^-19)r_0

r_0 = frac9 times 10^9 times 2 times 79 times (1.6 times 10^-19)^27.7 times 10^6 times 1.6 times 10^-19

r_0 = frac9 times 10^9 times 158 times 1.6 times 10^-197.7 times 10^6

r_0 = frac2275.2 times 10^-107.7 times 10^6 = 295.48 times 10^-16text m approx 2.95 times 10^-14text m

### Pattern Recognition Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy

k(Z_1e)(Z_2e)/r_0

. Alpha particle has charge

2e

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

More Atoms Previous-Year Questions — Page 4

Q36 jee_main_2024_30_jan_morning Bohr Model and Electron Energy

The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the

5^textth

excited state of a hydrogen atom is :

A. $4$ $4$
B. $frac14$ $\frac{1}{4}$
C. $frac12$ $\frac{1}{2}$
D. $1$ $1$

Solution

### Related Formula

textKE = -E

textPE = 2E

textKE = frac12 |textPE|

### Core Logic In any allowed Bohr orbit (for any value of principal quantum number

n

), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field.

|textPE| = 2 times textKE

### Step 1: Formulate the Ratio The question asks for the ratio of the magnitude of KE to the magnitude of PE.

fractextKE|textPE| = frac12

This ratio is independent of the orbit state. Even though it is the

5^textth

excited state, the ratio remains

frac12

. ### Pattern Recognition Energy relationships in Bohr orbits (and planetary motion):

K = -E = -U/2

. The state number (

n

) is given solely as a distractor. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Q53 jee_main_2024_30_jan_morning Hydrogen Energy Levels and Transitions

A electron of hydrogen atom on an excited state is having energy

E_n = -0.85 mathrm~eV

. The maximum number of allowed transitions to lower energy level is ....

Numerical Answer. Answer: 6 to 6

Solution

### Related Formula

E_n = -frac13.6n^2 mathrm~eV

textNumber of transitions = fracn(n - 1)2

### Core Logic First, identify the principal quantum number

n

corresponding to the energy

-0.85 mathrm~eV

. Then, use the combinatorics formula to find the total possible downward emission transitions. ### Step 1: Find Quantum State

E_n = -frac13.6n^2

-0.85 = -frac13.6n^2

n^2 = frac13.60.85 = 16

n = 4

### Step 2: Calculate Transitions Maximum number of transitions from

n=4

to lower levels (

n=3, 2, 1

= fracn(n - 1)2

= frac4(4 - 1)2 = frac122 = 6

### Pattern Recognition Energy states in Hydrogen are heavily standardized:

n=1 rightarrow -13.6

n=2 rightarrow -3.4

n=3 rightarrow -1.51

n=4 rightarrow -0.85

. Recognize

-0.85 mathrm~eV

as state 4 immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

Q36 jee_main_2024_31_jan_morning Hydrogen Spectrum

If the wavelength of the first member of Lyman series of hydrogen is

lambda

. The wavelength of the second member will be

A. $frac2732 lambda$ $\frac{27}{32} \lambda$
B. $frac3227lambda$ $\frac{32}{27}\lambda$
C. $frac275 lambda$ $\frac{27}{5} \lambda$
D. $frac527lambda$ $\frac{5}{27}\lambda$

Solution

### Related Formula

frac1lambda = R Z^2 left[ frac1n_1^2 - frac1n_2^2 right]

### Core Logic For the first member of the Lyman series of hydrogen (

n_1 = 1, n_2 = 2

frac1lambda = frac13.6 Z^2hc left[ frac11^2 - frac12^2 right]

frac1lambda = frac13.6 Z^2hc left[ frac34 right] dots dots (texti)

### Step 2: Second Member Calculation For the second member of the Lyman series (

n_1 = 1, n_2 = 3

frac1lambda' = frac13.6 Z^2hc left[ frac11^2 - frac13^2 right]

frac1lambda' = frac13.6 Z^2hc left[ frac89 right] dots dots (textii)

### Step 3: Ratio On dividing equation (i) by (ii):

fraclambda'lambda = frac3/48/9 = frac34 times frac98

fraclambda'lambda = frac2732

lambda' = frac2732 lambda

### Pattern Recognition Rydberg ratios between members of the same series are purely derived from the bracket terms

[1/n_1^2 - 1/n_2^2]

. For Lyman 1st and 2nd, the ratio is

(3/4) / (8/9) = 27/32

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and $frac14pi in_0 = 9 times 10^9$ in SI units)

Solution & Explanation

Reference Study Guides

More Atoms Previous-Year Questions — Page 4

Solution

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and frac14pi in_0 = 9 times 10^9$\frac{1}{4\pi \in_{0}} = 9 \times 10^{9}$ in SI units)

Solution & Explanation

Reference Study Guides

More Atoms Previous-Year Questions — Page 4

Solution

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If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and $frac14pi in_0 = 9 times 10^9$ in SI units)