Relativistic corrections become necessary when the expression for the kinetic energy 1/2 mv2, becomes comparable with mc2. where m is the mass of the At what de-Broglie wavelength, will relativistic corrections become important for an electron? (a) A=10nm (b) A =10-1 nm (c) A=10-4 nm (d) A=10-6 nm

This is a Multiple Choice Questions as classified in NCERT ExemplarAnswer-(c,d)Explanation- λ = h m v V= h/m λ V1= λ=10nm=10×10-9m=10-86.6×10-349×10-31×10-8=105m/sV1= λ=1/10nm=10-106.6×10-349×10-31×10-10=107m/sV1= λ=1/10000nm=10-136.6×10-349×10-31×10-13=1010m/sV1= λ=1/1000000nm=10-15

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Dual Nature of Radiation and Matter Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Relativistic corrections become necessary when the expression for the kinetic energy 1/2 mv², becomes comparable with mc². where m is the mass of the
At what de-Broglie wavelength, will relativistic corrections become important for an electron?
(a) A=10nm (b) A =10^-1 nm (c) A=10^-4nm (d) A=10^-6nm

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-(c,d)

Explanation- $λ = \frac{h}{m v}$

V= h/m $λ$

V1= $λ = 10 n m = 10 \times 10^{- 9} m = 10^{- 8}$

$\frac{6.6 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 8}} = 10^{5} m / s$

V1= $λ = 1 / 10 n m = 10^{- 10}$

$\frac{6.6 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 10}} = 10^{7} m / s$

V1= $λ = 1 / 10000 n m = 10^{- 13}$

$\frac{6.6 \times 10^{- 34}}{9 \times 10^{- 31} \times 10^{- 13}} = 10^{10} m / s$

V1= $λ = 1 / 1000000 n m = 10^{- 15}$

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Given below are two statements:

Statement I : Davission-Germer experiment establishes, the wave nature of electrons.

Statement II : If electrons have wave nature, they can interfere and show diffraction.

In the light of the above statements choose the correct answer from the option given below:

Raj Pandey

Based on theory

Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)² = 5.25]

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z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
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A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is: (Take hc = 1230 eV –nm).

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$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (i)

$_{}_{} \frac{}{}$ $K . E_{2} = 2 K . E_{1} = \frac{1230}{500} - ?$ - (ii)

from (i) & (ii)

$? = 0.615$ ev

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $υ_{1} .$ When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $υ_{2} .$ If $υ_{2} = x υ_{1},$ the value of x will be __________.

Vishal Baghel

hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2} m v_{1}^{2} = h υ_{0}$

$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

$v_{1} = \sqrt[]{\frac{2 h υ_{0}}{m}}$ - (1)

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

v₂ = $\sqrt[]{\frac{8 h υ_{0}}{m}}$ - (2)

$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

v₂ = 2v₁

Kindly consider the following

$f (x) = {\begin{matrix} \frac{2 x^{2} - 3 x + 2}{x - 2}, & x \neq 2 \\ 5, & x = 2 \end{matrix}$

alok kumar singh

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$

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