MHT CET 2021 20th September Morning Shift | Dual Nature of Radiation Question 77 | Physics

When a photosensitive surface is irradiated by light of wavelengths '$$\lambda_1$$' and '$$\lambda_2$$', kinetic energies of emitted photoelectrons are 'E$$_1$$' and 'E$$_2$$' respectively. The work function of photosensitive surface is

$$\frac{\left(\lambda_1 E_1-\lambda_2 E_2\right)}{\left(\lambda_2-\lambda_1\right)}$$

$$\frac{\left(\lambda_1 \mathrm{E}_1+\lambda_2 \mathrm{E}_2\right)}{\left(\lambda_2-\lambda_1\right)}$$

$$\frac{\left(\lambda_1 \mathrm{E}_2-\lambda_2 \mathrm{E}_1\right)}{\left(\lambda_2-\lambda_1\right)}$$

$$\frac{\left(\lambda_1 E_2+\lambda_2 E_1\right)}{\left(\lambda_2-\lambda_1\right)}$$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

The graph of stopping potential $V_s$ against frequency $v$ of incident radiation is plotted for two different metals $P$ and $Q$ as shown in the graph. $\phi_p$ and $\phi_Q$ are work-functions of $P$ and $Q$ respectively, then

MHT CET 2020 19th October Evening Shift Physics - Dual Nature of Radiation Question 39 English

$\phi_P>\phi_Q$

$\phi_P<\phi_Q$

$\phi_P=\phi_Q$

$\nu_0^{\prime}<\nu_0$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

If the maximum kinetic energy of emitted electrons in photoelectric effect is $3.2 \times 10^{-19} \mathrm{~J}$ and the work-function for metal is $6.63 \times 10^{-19} \mathrm{~J}$, then stopping potential and threshold wavelength respectively are

[Planck's constant, $h=6.63 \times 10^{34} \mathrm{~J}$-s]

[Velocity of light, $c=3 \times 10^8 \frac{\mathrm{~m}}{\mathrm{~s}}$ ]

[Charge on electron $=1.6 \times 10^{-19} \mathrm{C}$ ]

4V, 6000$\mathop A\limits^o$

3V, 4000$\mathop A\limits^o$

2V, 3000$\mathop A\limits^o$

1V, 1000$\mathop A\limits^o$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

The light of wavelength $$\lambda$$ incident on the surface of metal having work function $$\phi$$ emits the electrons. The maximum velocity of electrons emitted is [ $$c=$$ velocity of light, $$h=$$ Planck's constant, $$m=$$ mass of electron]

$$\left[\frac{2(h v-\phi) \lambda}{m c}\right]$$

$$\left[\frac{2(h c-\lambda \phi)}{m \lambda}\right]^{1 / 2}$$

$$\left[\frac{2(h c-\lambda)}{m \lambda}\right]^{1 / 2}$$

$$\left[\frac{2(h c-\phi)}{m \lambda}\right]$$

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