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In a hydrogen - like atom, an election is the ground state requires 476 ev to reach an excited level with quantum number 2n. If it makes a transition from this level to a lower level with quantum number n, it emits a photon of energy 40.8 eV. Find the atomic number Z of the element
  • a)
    1
  • b)
    3
  • c)
    6
  • d)
    10
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a hydrogen - like atom, an election is the ground state requires 47...
For hydrogen like atom 

Now. it requires 476 eV to reach 2n level from ground state 
E2n- E1 = 476
 ...(i)

It emits energy of 40.8 eV while making transition from 2n level to n level 
E2n - En = 40.8

 ...(ii)
Dividing (1) and (2)

n = 3
Putting n = 3 in (2) we get z = 6
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Most Upvoted Answer
In a hydrogen - like atom, an election is the ground state requires 47...
For hydrogen like atom 

Now. it requires 476 eV to reach 2n level from ground state 
E2n- E1 = 476
 ...(i)

It emits energy of 40.8 eV while making transition from 2n level to n level 
E2n - En = 40.8

 ...(ii)
Dividing (1) and (2)

n = 3
Putting n = 3 in (2) we get z = 6
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Community Answer
In a hydrogen - like atom, an election is the ground state requires 47...
To solve this question, we need to use the concept of energy levels in a hydrogen-like atom and the relationship between energy levels and quantum numbers.

First, let's understand the given information:
- The electron is in the ground state and needs 476 eV of energy to reach an excited level with quantum number 2n.
- The electron then undergoes a transition from this excited level to a lower level with quantum number n, emitting a photon of energy 40.8 eV.

Now, let's break down the solution into steps:

Step 1: Calculate the energy difference between the excited level and the lower level
The energy difference between two energy levels in an atom can be calculated using the formula:

ΔE = E2 - E1

where ΔE is the energy difference, E2 is the energy of the higher level, and E1 is the energy of the lower level.

In this case, the energy difference is given as 40.8 eV.

Step 2: Relate the energy difference to the quantum numbers
The energy levels in a hydrogen-like atom are given by the formula:

En = -13.6 eV / n^2

where En is the energy of the nth level and n is the principal quantum number.

We are given that the excited level has a quantum number of 2n, and the lower level has a quantum number of n. Therefore, we can write the energy difference as:

40.8 eV = -13.6 eV / (2n)^2 - (-13.6 eV / n^2)

Simplifying this equation gives:

40.8 eV = -13.6 eV * [(1 / (2n)^2) - (1 / n^2)]

Step 3: Solve for the value of n
To solve this equation, we substitute the given values and solve for n.

40.8 eV = -13.6 eV * [(1 / (2n)^2) - (1 / n^2)]

Dividing both sides by -13.6 eV and simplifying gives:

-3 = (1 / (2n)^2) - (1 / n^2)

Multiplying through by n^2(2n)^2 gives:

-3n^2(2n)^2 = (2n)^2 - n^2

Expanding and simplifying gives:

-3n^2(4n^2) = 4n^2 - n^2

-12n^4 = 3n^2

Dividing both sides by n^2 gives:

-12n^2 = 3

Simplifying further gives:

n^2 = -3/12

n^2 = -1/4

Since the square of a real number cannot be negative, we reject this solution.

Therefore, there is no solution for n, and hence no valid transition is possible.

Step 4: Conclusion
Since no valid transition is possible, the given information contradicts the laws of energy levels in a hydrogen-like atom. Therefore, the atomic number Z of the element cannot be determined from the given information.

Hence, the correct answer is option 'D' (not listed in the given options).
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In a hydrogen - like atom, an election is the ground state requires 476 ev to reach an excited level with quantum number 2n. If it makes a transition from this level to a lower level with quantum number n, it emits a photon of energy 40.8 eV. Find the atomic number Z of the elementa)1b)3c)6d)10Correct answer is option 'C'. Can you explain this answer?
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