>
Exams
>
Physics
>
thermal properties of matter
>
the minimum excitation energy of an electron revol
Question:
The minimum excitation energy of an electron revolving in the first orbit of hydrogen is:
Show Hint
Excitation energy is the energy required to move an electron from a lower to a higher energy level.
BITSAT - 2024
BITSAT
Updated On:
Mar 25, 2025
3.4 eV
8.5 eV
10.2 eV
13.6 eV
Hide Solution
Verified By Collegedunia
The Correct Option is
C
Solution and Explanation
Step 1:
{Energy Levels in the Hydrogen Atom}
The energy of an electron in the \( n^{th} \) orbit is: \[ E_n = \frac{-13.6}{n^2} { eV} \] For \( n = 1 \): \[ E_1 = -13.6 { eV} \] For \( n = 2 \): \[ E_2 = \frac{-13.6}{4} = -3.4 { eV} \]
Step 2:
{Calculating Excitation Energy}
\[ E_{{excitation}} = E_2 - E_1 \] \[ = (-3.4) - (-13.6) \] \[ = 10.2 { eV} \] Thus, the correct answer is \( 10.2 \) eV.
Download Solution in PDF
Was this answer helpful?
0
0
Top Questions on thermal properties of matter
In YDSE, monochromatic light falls on a screen 1.80 m from two slits separated by 2.08 mm. The first and second order bright fringes are separated by 0.553 mm. The wavelength of light used is:
BITSAT - 2024
Physics
thermal properties of matter
View Solution
If the distance between an object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be:
BITSAT - 2024
Physics
thermal properties of matter
View Solution
When a light ray incidents on the surface of a medium, the reflected ray is completely polarized. Then the angle between reflected and refracted rays is:
BITSAT - 2024
Physics
thermal properties of matter
View Solution
A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm. The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit will be:
BITSAT - 2024
Physics
thermal properties of matter
View Solution
Power of a biconvex lens is \( P \) diopter. When it is cut into two symmetrical halves by a plane containing the principal axis, the ratio of the power of two halves is:
BITSAT - 2024
Physics
thermal properties of matter
View Solution
View More Questions
Questions Asked in BITSAT exam
Let \( ABC \) be a triangle and \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of \( A, B, C \) respectively. Let \( D \) divide \( BC \) in the ratio \( 3:1 \) internally and \( E \) divide \( AD \) in the ratio \( 4:1 \) internally. Let \( BE \) meet \( AC \) in \( F \). If \( E \) divides \( BF \) in the ratio \( 3:2 \) internally then the position vector of \( F \) is:
BITSAT - 2024
Vectors
View Solution
Let \( \mathbf{a} = \hat{i} - \hat{k}, \mathbf{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}, \mathbf{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \). Then, \( [\mathbf{a} \, \mathbf{b} \, \mathbf{c}] \) depends on:}
BITSAT - 2024
Vectors
View Solution
Let the foot of perpendicular from a point \( P(1,2,-1) \) to the straight line \( L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1} \) be \( N \). Let a line be drawn from \( P \) parallel to the plane \( x + y + 2z = 0 \) which meets \( L \) at point \( Q \). If \( \alpha \) is the acute angle between the lines \( PN \) and \( PQ \), then \( \cos \alpha \) is equal to:
BITSAT - 2024
Plane
View Solution
The magnitude of projection of the line joining \( (3,4,5) \) and \( (4,6,3) \) on the line joining \( (-1,2,4) \) and \( (1,0,5) \) is:
BITSAT - 2024
Vectors
View Solution
The angle between the lines whose direction cosines are given by the equations \( 3l + m + 5n = 0 \) and \( 6m - 2n + 5l = 0 \) is:
BITSAT - 2024
Vectors
View Solution
View More Questions