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 Exemplar Answer-(c,d) Explanation- λ = h m v V= h/m λ V1= λ=10nm=10*10-9m=10-8 6.6*10-349*10-31*10-8=105m/s V1= λ=1/10nm=10-10 6.6*10-349*10-31*10-10=107m/s V1= λ=1/10000nm=10-13 6.6*10-349*10-31*10-13=1010m/s V1= λ=1/1000000nm=10-15

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

8 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= $λ = 10 n m = 10 * 10^{- 9} m = 10^{- 8}$

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

V1= $λ = 1 / 10 n m = 10^{- 10}$

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

V1= $λ = 1 / 10000 n m = 10^{- 13}$

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

V1= $λ = 1 / 1000000 n m = 10^{- 15}$

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Given below are two statements:

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

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hu = hu₀ + K.E

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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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