Physics Atoms

The Kα - X ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be _____ keV. (Round off to the nearest integer)
[h = 4.14 * 10⁻¹⁵ eVs, c = 3 * 10⁸ ms⁻¹]

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Contributor-Level 10

E_K - E_L = hc/λ_Kα
E_K - E_L = (4.14*10? ¹? * 3*10? )/0.071*10? = 17500 eV = 17.5 keV.
E_L = E_K - 17.5 = 27.5 - 17.5 = 10 keV.

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a month ago

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron?

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Contributor-Level 10

Energy of electron = 3 eV
It forms H atom in n=3 state. Energy released E = 3 - (-13.6/9) = 3 + 1.51 = 4.51 eV.
Photon Energy = 4.51 eV
Threshold energy = hc/λ = 12400eVÅ / 4000Å = 3.1 eV.
kE_max = 4.51 - 3.1 = 1.41 eV

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Consider the following statements:
(A) Atoms of each element emit characteristics spectrum.
(B) According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit.
(C) The density of nuclear matter depends on the size of the nucleus.
(D) A free neutron is stable but a free proton decay is possible.
(E) Radioactivity is an indication of the instability of nuclei.
Choose the correct answer from the options given below:

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Contributor-Level 10

Density of nucleus is constant.

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In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 Å. If the speed of electron is 2.2 * 10? m/s, then the current associated with the electron will be _______ * 10?² mA. [Take π as 22/7]

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Contributor-Level 10

r = 0.5Å = 0.5*10? ¹? m
v = 2.2*10? m/s
I =?
t = 2πr/v
I = e/t = ev/2πr = (1.6*10? ¹? * 2.2*10? )/ (2 * 22/7 * 0.5*10? ¹? )
I ≈ 1.12*10? ³ A = 1.12 mA = 112 * 10? ² mA

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a month ago

If 'f' denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0 (N?) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as: [λ is the radioactive decay constant]

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Contributor-Level 10

f = (N? - N? (e? )/N?
f = 1 - e?
df/dt = 0 - e? * -λ = λe?

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a month ago

In the given, figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E. The transitions A, B and C respectively represents:

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alok kumar singh

Contributor-Level 10

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

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a month ago

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be:

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Contributor-Level 10

$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{5^{2}}) = 13.6 * \frac{24}{25} e V$

$\Rightarrow \frac{h c}{λ} = 13.6 * \frac{24}{25} e V . . . . . . . . . . (1)$

With the help of conservation of linear momentum, we can write

$\frac{h}{λ} = m_{H} v_{H} \Rightarrow \frac{h c}{λ} = c m_{H} v_{H} \Rightarrow v_{H} = \frac{\frac{h c}{λ}}{c m_{H}} = \frac{13.6 * \frac{24}{25} * 1.6 * 1 0^{- 19}}{3 * 1 0^{8} * 1.67 * 1 0^{- 27}} = 4.17 m / s$

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a month ago

If $λ_{1} a n d λ_{2}$ are the wavelengths of the third member of Lyman and first member of Paschen series respectively, then the value of $λ_{1} : λ_{2} i s :$

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Payal Gupta

Contributor-Level 10

Lyman Series n_f = 1, ni = 2, 3.

Paschen series n_f = 3, ni = 4, 5, 6 -

$\frac{1}{λ_{1}} = R Z^{2} (\frac{1}{1^{2}} - \frac{1}{4^{2}})$

$\frac{1}{λ_{2}} = R Z^{2} (\frac{1}{3^{2}} - \frac{1}{4^{2}})$

$\frac{1 / λ_{1}}{1 / λ_{2}} = \frac{1 - \frac{1}{16}}{\frac{1}{9} - \frac{1}{16}} \Rightarrow \frac{λ_{2}}{λ_{1}} = \frac{\frac{15}{16}}{\frac{7}{9 * 16}} = \frac{15 * 9}{7} \Rightarrow \frac{λ_{1}}{λ_{2}} = \frac{7}{135}$

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a month ago

An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be:

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Contributor-Level 10

V = 1.24 * 10⁶ volt

$λ_{m i n} = ?$

$λ_{m i n} = \frac{h c}{e V} = \frac{1242 n m}{e (1.24 * 1 0^{6})} = \frac{1000 * 1 0^{- 9}}{1 0^{6}} \Rightarrow λ_{m i n} = 1 0^{- 3} n m$

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a month ago

According to Bohr atom model, in which of the following transitions will the frequency be maximum?

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