The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature…

Q: The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (kelvin) and mass m is :

(a) $\large \frac{h}{\sqrt{3 m k T}}$

(b) $\large \frac{2h}{\sqrt{3 m k T}}$

(c) $\large \frac{2h}{\sqrt{ m k T}}$

(d) $\large \frac{h}{\sqrt{ m k T}}$

Ans: (a)

Sol: As $\large E = \frac{3}{2}kT$

$\large \lambda = \frac{h}{\sqrt{2 m E}} $

$\large \lambda = \frac{h}{\sqrt{2 m (\frac{3}{2}kT) }} $

$\large = \frac{h}{\sqrt{3 m k T}}$

The Photoelectric threshold wavelength of silver is 3250 × 10-10 m . The velocity of the electron ejected…

Q: The Photoelectric threshold wavelength of silver is 3250 × 10-10 m . The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10-10 m is :

(a) 0.6 × 106 m/s

(b) 61 × 103 m/s

(c) 0.3 × 106 m/s

(a) 6 × 104 m/s

Ans: (a)

$\large \phi = \frac{1242}{325} = 3.82eV$

$\large E = \frac{1242}{253.6} = 4.89 eV$

E = φ + Kmax

Kmax = E – φ = 1.077 eV

$\large \frac{1}{2}m v_{max}^2 = 1.077 \times 1.6 \times 10^{-19} $

$\large v_{max} = (\frac{2 \times 1.077 \times 1.6 \times 10^{-19} }{9.1 \times 10^{-31}})^{1/2}$

vmax = 0.6 × 106 m/s

An electron of mass m with an initial velocity v = v0î (v0 > 0) enters an electric field E = -E0î (E0 = constant > 0) at t= 0 .

Q: An electron of mass m with an initial velocity v = v0î (v0 > 0) enters an electric field E = -E0î (E0 = constant > 0) at t= 0 . If λ0 is its De-Broglie wavelength initially , then its its de-Broglie wavelength at time t is

(a)λ0

(b)λ0

(c) λ0 (1 + eE0 t/m v0)

(c) λ0/ (1 + eE0 t/m v0)

Ans: (d)

Sol: Initial de-Broglie wavelength λ0 = h/mv0

Wavelength at time t , λ = h/mv

where , v = v0 + eE0t/m

λ/λ0 = v0/v

λ/λ0 = v0/(v0 + eE0t/m)

λ/λ0 = v0/v0(1 + eE0t/mv0)

λ = λ0/ (1 + eE0 t/m v0)

when the light of frequency 2ν0 (Where ν0 is threshold frequency ) is incident on a metal plate , the maximum velocity

Q: when the light of frequency 2ν0 (Where ν0 is threshold frequency ) is incident on a metal plate , the maximum velocity of electron emitted is v1 . When the frequency of incident radiation is increased to 5ν0 , the maximum velocity of electron emitted from the same plate is v2 . The ratio of v1 to v2

(a)4:1

(b)2:1

(c)1:4

(d)1:2

Ans: (d)

Sol: In case (i)

h.2ν0 = h.ν0 + (1/2)m v12

h.ν0 = (1/2)m v12 …(i)

In case (ii)

h.5ν0 = h.ν0 + (1/2)m v22

h.4ν0 = (1/2)m v22 …(ii)

Dividing equations (i) and (ii) we get

v1/v2 = 1/2

From the above figure the values of stopping potentials for M1 and M2 for a frequency v3 (> vo2) ….

Q: From the above figure the values of stopping potentials for M1 and M2 for a frequency v3 (> v02) of the incident radiations are V1 and V2 respectively. Then the slope of the line is equal to

Numerical

(a) $ \displaystyle \frac{V_2 -V_1}{\nu_{02} – \nu_{01}} $

(b) $ \displaystyle \frac{V_1 -V_2}{\nu_{02} – \nu_{01}} $

(c) $ \displaystyle \frac{V_2}{\nu_{02} – \nu_{01}} $

(d) $ \displaystyle \frac{V_1}{\nu_{02} – \nu_{01}} $

Ans: (a)

Sol:
$ \displaystyle h\nu_{01} = \phi + eV_1 $

$ \displaystyle V_1 = \frac{h \nu_{01}}{e} -\frac{\phi}{e} $ …(i)

Similarly ,

$ \displaystyle V_2 = \frac{h \nu_{02}}{e} -\frac{\phi}{e} $ …(ii)

subtracting (i) from(ii)

$ \displaystyle V_2 – V_1 = \frac{h}{e}(\nu_{02}-\nu_{01}) $

$ \displaystyle \frac{h}{e} = \frac{V_2 -V_1}{\nu_{02} – \nu_{01}} $