An \alpha -particle and a proton are accelerated from rest by a potential difference of 100 V . After this, their de Broglie wavelengths are \lambda_{\alpha} and \lambda_{p} respectively. The ratio \dfrac{\lambda_{p}}{\lambda_{\alpha}} , to the nearest integer, is:

We have, \dfrac{1}{2}mv^{2}=eV De broglie wavelength= \lambda=\dfrac{h}{mv} \lambda = \dfrac{h}{\sqrt{2meV}} \lambda_{p}=\dfrac{h}{\sqrt{2(m)eV}} ....(1) \lambda_{\alpha}=\dfrac{h}{\sqrt{2(4m)2eV}} ....(2)Now, \dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{8} \simeq 3

Dual Nature Of Radiation And Matter

Q: A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is : (take the proton mass, m_p = (5/3) \times 10^{-27}kg and h/e = 4.2\times 10^{-15}J.s/C; \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 m/F; 1 fm = 10^{-15}m )

Given here, q_{1} = 120 \times e q_{2} = e ( charge on proton) It is based on conservation of energy of proton. \dfrac{1}{2}mv^2 = k \dfrac{q_1q_2}{R} v = \sqrt{\dfrac{2Kq_1q_2}{mR}} Also, we know, de’broglie wavelength, \lambda =\dfrac{h}{mv} \lambda = \dfrac{h\sqrt{R}}{\sqrt{2mKq_1q_2}}\\ Thus, \lambda = \dfrac{h\sqrt{10\times10^{-15}}}{\sqrt{2\times\dfrac{5}{3}\times10^{-27}\times9\times10^9\times120\times e^2}}\\ \lambda =\dfrac{\dfrac{h}{e}\times10^{-7}}{6\times10^{-8}}\\ \lambda =\dfrac{4.2\times10^{-15}\times10^{-7}}{6\times10^{-8}}\\ \lambda =7\times10^{-15}\ m = 7\ fm \\ Option A is correct.

Q: A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3\times 10^{8}m/s . The final momentum of the object is:

Sol . (B) t=100\times 10^{-9} sec,\ P=30\times 10^{-3} Watt, C=3\times 10^{8}m/s Incident energy = Pt By debroglie wavelength, \lambda=h/p and Energy of photon is E=hc/\lambda From above two equations, E=pc Momentum p=\dfrac{Pt}{C}=\dfrac{30\times 10^{-3}\times 100\times 10^{-9}}{3\times 10^{8}}=1.0\times 10^{-17} kg ms ^{-1} So, the answer is option (B).

Q: The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is

Slope of graph is h/e = constant \Rightarrow 1

Q: A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm . The maximum speeds of the photoelectrons corresponding to these wavelengths are u_1 and u_2 , respectively. If the ratio u_1 : u_2 = 2 : 1 and hc = 1240 eV nm , the work function of the metal is nearly :

On solving we get \displaystyle \dfrac{\dfrac{hc}{\lambda_1}- \phi}{\dfrac{hc}{\lambda_2} - \phi} = \dfrac{u_1^2}{u_2^1} On putting the values we get \phi = 3.7 eV

Q: The graph between 1/\lambda and stopping potential (V) of three metals having work functions \phi_{1},\ \phi_{2} and \phi_{3} in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) \mathrm{i}\mathrm{s}/\mathrm{a}\mathrm{r}\mathrm{e} correct? [Here \lambda is the wavelength of the incident ray].

From photo electric effect, eV=\dfrac{hc}{\lambda}-\phi where V= stopping potential and \phi= work function. Slope , m=\tan\theta=hc/e (compare above equation with y=mx-c )If V=0 \Rightarrow \dfrac{\phi}{hc}=\dfrac{1}{\lambda} Now from given graph,for metal 1 : \dfrac{\phi_1}{hc}=0.001 for metal 2: \dfrac{\phi_2}{hc}=0.002 for metal 3: \dfrac{\phi_3}{hc}=0.004 Thus, \phi_1:\phi_2:\phi_3=1:2:4 Also, \phi=\dfrac{hc}{\lambda_0} where \lambda_0 be the threshold wavelength. For metal 2 : \lambda_{02}=1/0.002=500 nm For metal 3 : \lambda_{02}=1/0.004=250 nm As the wavelength of violet colour is around 400 nm so only metal 2 can eject violet colour light but not metal 3. Thus, options A and C is right.

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Dual Nature Of Radiation And Matter

- Wondering what kind of questions that can be asked in next exam? Here you go!

JEE Advanced

Single Correct

A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is :

(take the proton mass,

m_{p} = (5 / 3) \times 1 0^{- 27} k g

and

h / e = 4.2 \times 1 0^{- 15} J . s / C; \frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} m / F; 1 f m = 1 0^{- 15} m

)

7 f m

8 f m

9 f m

10 f m

View solution

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30

m W

and the speed of light is

3 \times 1 0^{8} m / s

. The final momentum of the object is:

0.3 \times 1 0^{- 17} k g m s^{- 1}

1.0 \times 1 0^{- 17} k g m s^{- 1}

3.0 \times 1 0^{- 17} k g m s^{- 1}

9.0 \times 1 0^{- 17} k g m s^{- 1}

View solution

The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is

0.33

0.67

1.33

View solution

A metal surface is illuminated by light of two different wavelengths

248 n m

and

310 n m

. The maximum speeds of the photoelectrons corresponding to these wavelengths are

u_{1}

and

u_{2}

, respectively. If the ratio

u_{1} : u_{2} = 2 : 1

and

h c = 1240 e V n m

, the work function of the metal is nearly :

3.7 e V

3.2 e V

2.8 e V

2.5 e V

View solution

Multiple Correct

The graph between

1 / λ

and stopping potential (V) of three metals having work functions

ϕ_{1}, ϕ_{2}

and

ϕ_{3}

in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s)

i s / a r e

correct? [Here

λ

is the wavelength of the incident ray].

This question has multiple correct options

Ratio of work functions

ϕ_{1}

ϕ_{2}

ϕ_{3} = 1

: 2 : 4

Ratio of work functions

ϕ_{1}

ϕ_{2}

ϕ_{3} = 4

: 2 : 1

tan Θ

is directly proportional to

h c / e

, where

h

is Planck's constant and

c

is the speed of light.

The violet colour light can eject photoelectrons from metals 2 and 3.

View solution

Numerical Type I

α

-particle and a proton are accelerated from rest by a potential difference of

100 V

. After this, their de Broglie wavelengths are

λ_{α}

and

λ_{p}

respectively. The ratio

\frac{λ _{p}}{λ _{α}}

, to the nearest integer, is:

View solution